MATH 409 Lecture 17

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Lecture Slides

Examples

$\mathrm {e} ^{x}>x+1$ for all $x\neq 0$.

Consider $f(x)=\mathrm {e} ^{x}-x-1$ for $x\in \mathbb {R}$. Our goal is to show that $f$ is greater than $0$ for all $x\neq 0$. This function is differentiable on $\mathbb {R}$ and $f'(x)=\mathrm {e} ^{x}-1$ for all $x\in \mathbb {R}$. We observe that $f'$ is strictly increasing. Since $f'(0)=0$, we have $f'(x)<0$ for all $x<0$ and $f'(x)>0$ for all $x>0$.

It follows that $f$ is strictly decreasing on $(-\infty ,0]$ and strictly increasing on $[0,\infty )$. As a consequence, $f(x)>f(0)=0$ for all $x\neq 0$. Thus $\mathrm {e} ^{x}>x+1$ for $x\neq 0$.

$\log {x}<x-1$ for all $x>0$, $x\neq 1$.

By above, $\mathrm {e} ^{x-1}>(x-1)+1=x$ for all $x\neq 1$. Since the natural logarithm is strictly increasing on $(0,\infty )$, it follows that $\log {\mathrm {e} ^{x-1}}>\log {x}$ for $x>0$, $x\neq 1$. Equivalently $x-1>\log {x}$ for $x>0$, $x\neq 1$.

Bernoulli's Inequality. $\left(1-x\right)^{\alpha }>1-\alpha \,x$ for all $x\in (0,1)$ and $\alpha >1$.

Fix an arbitrary $\alpha >1$ and consider

$f(x)=(1-x)^{\alpha }-1+\alpha \,x$

This function is differentiable on $[0,1)$ and $f'(x)=-\alpha \,\left(1-x\right)^{\alpha -1}+\alpha$ for all $x\in [0,1)$. Since $\alpha -1>0$, we obtain that $\left(1-x\right)^{\alpha -1}<1$ for $x\in (0,1)$. Hence $f'(x)>0$ for $x\in (0,1)$.It follows that the function is strictly increasing on $[0,1)$. As a consequence, $f(x)>f(0)=0$ for all $x\in (0,1)$.

$\left(1-x\right)^{\alpha }<1-\alpha \,x+{\frac {\alpha \,\left(\alpha -1\right)}{2}}\,x^{2}$ for all $x\in (0,1)$ and $\alpha >2$.

Let us fix an arbitrary $\alpha >2$ and consider a function

$f(x)=\left(1-x\right)^{\alpha }-1+\alpha \,x-{\frac {1}{2}}\,\alpha \,\left(\alpha -1\right)\,x^{2}$

This function is infinitely differentiable on $[0,1)$ and for all $x\in [0,1)$,

${\begin{aligned}f'(x)&=-\alpha \,(1-x)^{\alpha -1}+\alpha -\alpha \,\left(\alpha -1\right)\,x\\f''(x)&=\alpha \,\left(\alpha -1\right)\,\left(1-x\right)^{\alpha -2}-\alpha \,\left(\alpha -1\right)\end{aligned}}$

Since $\alpha -2>0$, we obtain that $f''<0$ for $x\in (0,1)$. It follows that the derivative $f'$ is strictly decreasing on Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [0,1)} . As a consequence, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f'(x) < f'(0) = 0} for all $x\in (0,1)$. Now it follows that the function is strictly decreasing on Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [0,1)} . Consequently Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x) < f(0) = 0} for all $x\in (0,1)$. The required inequality follows.

The function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x) = \left( 1+x \right)^{\frac{1}{x}}} is strictly decreasing on Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (0,\infty)} .

(The limit at 0 is $\mathrm {e}$)

Consider a function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g(x) = \log{f(x)}} for $x>0$. For every Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x > 0} , we have Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g(x) = \frac{\log{\left( 1+x \right)}}{x}} . This function is differentiable on Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (0,\infty)} , and therefore the original function $f$ is differentiable (it is the exponentiation of this function):

$g'(x)={\frac {{\frac {1}{1+x}}-\log {\left(1+x\right)}}{x^{2}}}$

Now we introduce another function $h(x)={\frac {x}{1+x}}-\log {\left(1+x\right)}=1-{\frac {1}{1+x}}-\log {\left(1+x\right)}$, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \ge 0} .

Notice that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h(x) = x^2 \, g'(x)} for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x > 0} . The function $h$ is differentiable on Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [0, \infty)} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h'(x) = \frac{1}{(1+x)^2} - \frac{1}{1+x} < 0} for all Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x > 0} . It follows that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h} is strictly decreasing on $[0,\infty )$. In particular, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h(x) < h(0) = 0} for $x>0$. Then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g'(x) < 0} for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x > 0} as well. Therefore $g$ is strictly decreasing on Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (0, \infty)} . Since Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} is the composition of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g} with the strictly increasing function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y(x) = \mathrm{e}^{x}} , it is also strictly decreasing on $(0,\infty )$.

Taylor's Formula

Theorem. If a function $f:I\to \mathbb {R}$ is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n + 1} times differentiable on an open interval Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I} , then for any two points Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x, x_0 \in I} , there is a point Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c} between Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_0} such that

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x) = f(x_0) + \sum_{k=1}^n \frac{f^{(k)}(x_0)}{k!} \, \left( x - x_0 \right)^k + \frac{f^{(n+1)}(c)}{\left( n + 1 \right)!} \, \left( x - x_0 \right)^{n+1}}

This function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P_n^{f,x_0}} is called the Taylor polynomial of order $n$ generated by Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} centered at $x_{0}$.

It provides information on the remainder term $r_{n}^{f,x_{0}}=f-P_{n}^{f,x_{0}}$. In many cases, this information allows us to estimate $\left|r_{n}^{f,x_{0}}(x)\right|$, the error in the estimate, or to prove an inequality of the form $f(x)<P_{n}^{f,x_{0}}(x)$ or $f(x)>P_{n}^{f,x_{0}}(x)$.

This will come in handy on the homework

l'Hôspital's Rule

(May also be spelled l'Hôpital's Rule) Helps us to compute limit of quotients in those cases where limit theorems do not apply due to indeterminacy of the form ${\frac {0}{0}}$ or ${\frac {\infty }{\infty }}$.

Theorem. Let $a$ be an extended real number. Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I} be an open interval such that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a \in I} or Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a} is an endpoint of $I$. Suppose that $f$ and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g} are differentiable on Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I} and that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g(x), g'(x) \ne 0} for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \in I \setminus \left\{ a \right\}} . Suppose further that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \underset{x \in I}{\lim_{x \to a}} f(x) = \underset{x \in I}{\lim_{x \to a}} g(x) = A}

Where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A = 0 \mbox{ or } \infty} , If the limit Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \underset{x \in I}{\lim_{x \to a}} \frac{f'(x)}{g'(x)}} existst (finite or infinite), then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \underset{x \in I}{\lim_{x \to a}} \frac{f(x)}{g(x)} = \underset{x \in I}{\lim_{x \to a}} \frac{f'(x)}{g'(x)}} .

Proof. (in case $\lim _{x\to a^{+}}{\frac {0}{0}}$) We extend Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g} to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I \cup \left\{ a \right\}} by letting Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(a) = g(a) = 0} . By hypothesis, $f$ and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g} are continuous on Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I \cup \left\{ a \right\}} and differentiable on Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I} . By the Generalized Mean Value Theorem, for any Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \in I} , there exists $c_{x}\in (a,x)$ such that

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g'(c_x) \, \left( f(x)-f(a) \right) = f'(c_x) \, \left( g(x) - g(a) \right)}

That is, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g'(c_x) \, f(x) = f'(c_x) \, g(x)} . Since Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g(c_x), g'(c_x) \ne 0} , we obtain Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{f(x)}{g(x)} = \frac{f'(c_x)}{g'(c_x)}} . Since $c_{x}\in (a,x)$, we have Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c_x \to a} as Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \to a^+} . It follows that

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x \to a^+} \frac{f(x)}{g(x)} = \lim_{x \to a^+} \frac{f'(c_x)}{g'(c_x)} = \lim_{c \to a^+} \frac{f'(c)}{g'(c)}}

The theorem includes several similar rules corresponding to various kinds of limits ($x\to a^{+}$, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \to a^-} , $x\to a$ for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a \in \mathbb{R}} , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \to +\infty} , $x\to -\infty$), and the two types of indeterminacy (Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{0}{0}} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\infty}{\infty}} )

quod erat demonstrandum

Examples

$\lim _{x\to 0}{\frac {1+\cos {x}}{x^{2}}}$

The functions Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x) = 1 - \cos{x}} and $g(x)=x^{2}$ are infinitely differentiable on Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbb{R}} . We have Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x \to 0} f(x) = f(0) = 0} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x \to 0} g(x) = g(0) = 0} .

Further, $f'(x)=\sin {x}$ and $g'(x)=2x$. We obtain that the form is still indeterminate at $0$.

Even further, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f''(x) = \cos{x}} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g''(x) = 2} . We obtain that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x \to 0} f''(x) = f''(0) = 1} and $\lim _{x\to 0}g''(x)=g''(0)=2$. It follows that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x \to 0} \frac{f''(x)}{g''(0)} = \frac{1}{2}} .

By l'Hôspital's Rule, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x \to 0} \frac{f'(x)}{g'(x)} = \frac{1}{2}} , then by extension Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x \to 0} \frac{f(x)}{g(x)} = \frac{1}{2}} .