mathematics-class-12-unit-06-chapter-12-derivatives-l-12-12-rdxxihmjcnu

### <Success style="font-size:25px"> Derivatives L-12 </Success>
### <primary style="font-size:24px"> L'Hopital's rule </primary>
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- This is used to calculate limits of the form $\lim _{x \rightarrow c} \frac{f(-x)}{g(x)}$, where $c$ is an extended real number, ie. $c$ is either a real number or $\pm \infty$.

- Special case: Suppose $f(x)$ and $g(x)$ are continuously differentiable functions in some interval I containing $C$.

- Also, assume that $f(c)=g(c)=0$.
- And, $ ~ g^{\prime}(c) \neq 0$.

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<footer style = "font-size: 18px"> $\rightarrow$ Derivatives lecture-12 $\rightarrow$ <span style = "color:blue"> L'Hopital's rule </span> $\rightarrow$ L'Hopital's rule $\rightarrow$ L'Hopital's rule </footer>

### <Success style="font-size:25px"> Derivatives L-12 </Success>
### <primary style="font-size:24px"> L'Hopital's rule </primary>
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- Now,

- $\lim _{x \rightarrow c} \frac{f(x)-f(c)}{x-c}=f^{\prime}(c)$

- $\begin{aligned}& \text { and } \lim _{x \rightarrow c} \frac{g(x)-g(c)}{x-c}=g^{\prime}(c) \neq 0 \\\\\therefore \quad & \lim _{x \rightarrow c} \frac{\left(\frac{f(x)-f(c)}{x-c}\right)}{\frac{g(x)-g(c)}{x-c}}=\frac{f^{\prime}(c)}{g^{\prime}(c)}\end{aligned}$

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<footer style = "font-size: 18px">L'Hopital's rule $\rightarrow$ L'Hopital's rule $\rightarrow$ <span style = "color:blue"> L'Hopital's rule </span> $\rightarrow$ L'Hopital's rule $\rightarrow$ L'Hopital's rule </footer>

### <Success style="font-size:25px"> Derivatives L-12 </Success>
### <primary style="font-size:24px"> L'Hopital's rule </primary>
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- $\therefore \quad \lim _{x \rightarrow c} \frac{f(x)}{g(x)}=\frac{f^{\prime}(c)}{g^{\prime}(c)}=\frac{\lim _{x \rightarrow c} f^{\prime}(x)}{\lim _{x \rightarrow c} g^{\prime}(x)}$

- because $f^{\prime}(x)$ \& $g^{\prime}(x)$ are assumed to be continuous at $x=c$.

- Hence, $\lim _{x \rightarrow c} \frac{f(x)}{g(x)}=\lim _{x \rightarrow c} \frac{f^{\prime}(x)}{g^{\prime}(x)}$

- The above rule, which is valid for more general cases, is known as the L'Hôpital's rule.

- L'Hopital's : French mathematician (pronounced is lopital)

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### <Success style="font-size:25px"> Derivatives L-12 </Success>
### <primary style="font-size:24px"> L'Hopital's rule </primary>
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- Suppose $\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c} g(x)=0$ or $\pm \infty$.

- $\left(i.e. \lim _{x \rightarrow c} \frac{f(x)}{g(x)} \text{is of the form} \frac{0}{0} \text{or} \frac{\infty}{\infty} \right)$

- Also, assume that $\lim _{x \rightarrow c} \frac{f^{\prime}(x)}{g^{\prime}(x)}$ exists and $g^{\prime}(x) \neq 0 \quad \forall x$ in the interval I except possibly at $x=c$.

- Then, $\lim _{x \rightarrow c} \frac{f(x)}{g(x)}$ exist and

- $\lim _{x \rightarrow c} \frac{f(x)}{g(x)}=\lim _{x \rightarrow c} \frac{f^{\prime}(x)}{g^{\prime}(x)} \text {. }$
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### <Success style="font-size:25px"> Derivatives L-12 </Success>
### <primary style="font-size:24px"> Problem-1 </primary>
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- $\lim _{x \rightarrow 0} \frac{\sin x}{x} \quad\left(\frac{0}{0}\right.$ form)

- Solution:

- $\lim _{x \rightarrow 0} \frac{\frac{d}{dx} \sin x}{\frac{d}{d x} x}=\lim _{x \rightarrow 0} \frac{\cos x}{1}=\frac{\cos 0}{1}=1 \text {. }$

- By the L'Hopital's rule,

- $\lim _{x \rightarrow 0} \frac{\sin x}{x}=1 \text {. }$

- Note that here we used the fact that
- $\frac{d}{d x}(\sin x)=\cos x \text {. }$

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<footer style = "font-size: 18px">L'Hopital's rule $\rightarrow$ L'Hopital's rule $\rightarrow$ <span style = "color:blue"> Problem-1 </span> $\rightarrow$ Problem-2 $\rightarrow$ Some complications </footer>

### <Success style="font-size:25px"> Derivatives L-12 </Success>
### <primary style="font-size:24px"> Problem-2 </primary>
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- $\begin{aligned}& \lim _{x \rightarrow 0} \frac{e^x-1-x}{x^2} \quad \text { ( } \frac{0}{0} \text { form) } \\\\& \quad \\\\& \text{Solution:} \\\\& \quad \\\\& =\lim _{x \rightarrow 0} \frac{e^x-1}{2 x} \quad \text { (by L'Höpital) } \\\\& =\lim _{x \rightarrow 0} \frac{e^x}{2} \quad \text { (by L'Hopital's) }\\\\& =\frac{e^0}{2}=\frac{1}{2} .\end{aligned} $

- Remark: We might have to apply L'Hópitals rule several times in order to calculate the limit.

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<footer style = "font-size: 18px">L'Hopital's rule $\rightarrow$ Problem-1 $\rightarrow$ <span style = "color:blue"> Problem-2 </span> $\rightarrow$ Some complications $\rightarrow$ Example </footer>

### <Success style="font-size:25px"> Derivatives L-12 </Success>
### <primary style="font-size:24px"> Some complications </primary>
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- $ \lim _{x \rightarrow \infty} \frac{e^x+e^{-x}}{e^{-x}-e^{-x}} \quad\left(\frac{\infty}{\infty}\right. \text { form) } $

- $ L^{\prime} H=\lim _{x \rightarrow \infty} \frac{e^x-e^{-x}}{e^x+e^{-x}} \quad \text { ( } \frac{\infty}{\infty} \text { form) } $

- $ L^{\prime} H = \lim _{x \rightarrow \infty} \frac{e^x+e^{-x}}{e^x-e^{-x}} \text { which is the original limit itself. } $

- So we wont be able to calculate the limit by applying L'Hopital's rule directly.

- However, if we put $e^x=y$, the as $x \rightarrow \infty, y \rightarrow \infty$.

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<footer style = "font-size: 18px">Problem-1 $\rightarrow$ Problem-2 $\rightarrow$ <span style = "color:blue"> Some complications </span> $\rightarrow$ Example $\rightarrow$ Example </footer>

### <Success style="font-size:25px"> Derivatives L-12 </Success>
### <primary style="font-size:24px"> Example </primary>
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- $\begin{aligned}& \quad \frac{e^x+e^{-x}}{e^{-x}-e^{-x}}=\frac{y+\frac{1}{y}}{y-\frac{1}{y}}=\frac{y^2+1}{y^2-1} \\\\& \therefore \quad \lim _{x \rightarrow \infty} \frac{e^{-x}+e^{-x}}{e^x-e^{-x}}=\lim _{y \rightarrow \infty} \frac{y^2+1}{y^2-1}=\lim _{y \rightarrow \infty} \frac{1+\frac{1}{y^2}}{1-\frac{1}{y}} \\\\&=\frac{1+0}{1-0}=1 .\end{aligned} $

- Or, we could use L'Hopital's  rule to write

- $\lim _{y \rightarrow \infty} \frac{y^2+1}{y^2-1} \stackrel{L^{\prime} H}{=} \lim _{y \rightarrow \infty} \frac{2 y}{2 y}=1 \text {. } $

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<footer style = "font-size: 18px">Problem-2 $\rightarrow$ Some complications $\rightarrow$ <span style = "color:blue"> Example </span> $\rightarrow$ Example $\rightarrow$ $\infty \cdot 0$ form </footer>

### <Success style="font-size:25px"> Derivatives L-12 </Success>
### <primary style="font-size:24px"> Example </primary>
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- $\begin{aligned}& \lim _{x \rightarrow \infty} \frac{\sqrt{x}+\frac{1}{\sqrt{x}}}{\sqrt{x}-\frac{1}{\sqrt{x}}} \quad\left(\frac{\infty}{\infty}\right. \text { form) } \\\\& \stackrel{L^{\prime} H}{=} \lim _{x \rightarrow \infty} \frac{\frac{1}{2 \sqrt{x}}+\frac{-1}{2} x^{-3 / 2}}{\frac{1}{2 \sqrt{x}}+\frac{1}{2} x^{-3 / 2}} \quad (\frac{0}{0}\text{from})\\\\ & \lim _{x \rightarrow \infty} \frac{-\frac{1}{4} x^{-3/2} +\frac{3}{4} x^{-5/ 2}}{-\frac{1}{4} x^{-3/2}-\frac{3}{4} x^{-5 / 2}} \quad (\frac{0}{0} \text { form) }) \\\\&\end{aligned} $

- =... becomes more \& more complicated.

- However, $\begin{aligned} \lim _{x \rightarrow \infty} \frac{\sqrt{x}+\frac{1}{\sqrt{x}}}{\sqrt{x}-\frac{1}{\sqrt{x}}}=\lim _{x \rightarrow \infty} \frac{x+1}{x-1} & \stackrel{L'H}{=} \lim _{x \rightarrow \infty} \frac{1}{1}  =1 .\end{aligned}$

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<footer style = "font-size: 18px">Some complications $\rightarrow$ Example $\rightarrow$ <span style = "color:blue"> Example </span> $\rightarrow$ $\infty \cdot 0$ form $\rightarrow$ $\frac{\infty}{\infty}$ form </footer>

### <Success style="font-size:25px"> Derivatives L-12 </Success>
### <primary style="font-size:24px"> $\infty \cdot 0$ form </primary>
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- L'Hôpital's Rule can be used for other indeterminate forms like

- $0 . \infty, \infty-\infty$, $1^{\infty}, 0^{\circ}$, etc. by somehow changing into $\frac{0}{0}$ a $\frac{\infty}{\infty}$ form.

- Examples: (1) $\lim _{x \rightarrow \infty} x^2 e^{-x} \quad$ ( $\infty .0$ form)

- $\begin{aligned}& =\lim _{x \rightarrow \infty} \frac{x^2}{e^{x}} \quad \text { ( } \frac{\infty}{\infty} \text { form) } \\\\&  \stackrel{L^{\prime} H}{=}  \lim _{x \rightarrow \infty} \frac{2 x}{e^x} \quad \text { ( } \frac{\infty}{\infty} \text { form) } \\\\& \stackrel{L^{\prime} H}{=} \lim _{x \rightarrow \infty} \frac{2}{e^x}=0 \text {. } \\& \end{aligned} $

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<footer style = "font-size: 18px">Example $\rightarrow$ Example $\rightarrow$ <span style = "color:blue"> $\infty \cdot 0$ form </span> $\rightarrow$ $\frac{\infty}{\infty}$ form $\rightarrow$ $\infty-\infty $ form </footer>

### <Success style="font-size:25px"> Derivatives L-12 </Success>
### <primary style="font-size:24px"> $\frac{\infty}{\infty}$ form </primary>
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- More generally, we can show that

- $\lim _{x \rightarrow \infty} x^n \cdot e^{-x}=0 \text { for any positive integer n }$

- $\lim _{x \rightarrow 0^{+}} x \ln x $ 
 
- (here we are taking $\lim _{x \rightarrow 0^{+}} ~ $ because $\ln x$ is defined  only for $x>0$)
 
- As $x \rightarrow 0+, \ln x \rightarrow-\infty \quad$  $y=\ln x$

- Let's write $x \ln x=\frac{\ln x}{x^{-1}}$ ( $\frac{-\infty}{\infty}$ form $y$)

- $\lim _{x \rightarrow 0^{+}} x \ln x  =\lim _{x \rightarrow 0^{+}} \frac{\ln x}{x^{-1}} \\\\ \stackrel{L^{\prime} H}{=}\lim _{x \rightarrow 0^{+}} \frac{1 / x}{-1 / x^2}=\lim _{x \rightarrow 0+}(-x)=0 .$

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<footer style = "font-size: 18px">Example $\rightarrow$ $\infty \cdot 0$ form $\rightarrow$ <span style = "color:blue"> $\frac{\infty}{\infty}$ form </span> $\rightarrow$ $\infty-\infty $ form $\rightarrow$ $\frac{0}{0}$ form </footer>

### <Success style="font-size:25px"> Derivatives L-12 </Success>
### <primary style="font-size:24px"> $\infty-\infty $ form </primary>
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- $\begin{aligned}& \lim _{x \rightarrow 0}\left(\frac{1}{x}-\frac{1}{\sin x}\right) \quad(\infty-\infty \text { form }) \\\\& =\lim _{x \rightarrow 0} \frac{\sin x-x}{x \sin x} quad\left(\frac{0}{0} \text { form }\right) \\\\&\stackrel{L^{\prime} H}{=} \lim _{x \rightarrow 0} \frac{\cos x-1}{\sin x+x \cos x} \quad\left(\frac{0}{0} \text { form }\right) \\\\&\stackrel{L^{\prime} H}{=}\lim _{x \rightarrow 0} \frac{-\sin x}{\cos x+\cos x-x \sin x} =\lim _{x \rightarrow 0} \frac{-\sin x}{2 \sin x-x \sin x}=\frac{0}{2}=0 .\end{aligned} $

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<footer style = "font-size: 18px">$\infty \cdot 0$ form $\rightarrow$ $\frac{\infty}{\infty}$ form $\rightarrow$ <span style = "color:blue"> $\infty-\infty $ form </span> $\rightarrow$ $\frac{0}{0}$ form $\rightarrow$ $ 0^0 $ form </footer>

### <Success style="font-size:25px"> Derivatives L-12 </Success>
### <primary style="font-size:24px"> $\frac{0}{0}$ form </primary>
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- $\begin{aligned}& \lim _{x \rightarrow 1^{+}}(x-1) \tan \left(\frac{\pi}{2} x\right)(0,-\infty) \\\\& =\lim _{x \rightarrow 1} \frac{x-1}{\cot \left(\frac{\pi}{2} x\right)} \quad\left(\frac{0}{0} \text { form }\right) \\\\&\stackrel{L^{\prime} H}{=}\lim _{x \rightarrow 1+} \frac{1}{-\operatorname{cosec}^2\left(\frac{\pi}{2} x\right) \cdot \frac{\pi}{2}} \\\\& =\lim _{x \rightarrow 1}-\frac{2}{\pi} \sin ^2\left(\frac{\pi}{2} x\right)=\frac{-2}{\pi} \sin ^2\left(\frac{\pi}{2}\right)=-\frac{2}{\pi} \text {. } \\\\&\end{aligned} $

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<footer style = "font-size: 18px">$\frac{\infty}{\infty}$ form $\rightarrow$ $\infty-\infty $ form $\rightarrow$ <span style = "color:blue"> $\frac{0}{0}$ form </span> $\rightarrow$ $ 0^0 $ form $\rightarrow$ Remark </footer>

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### <primary style="font-size:24px"> $ 0^0 $ form </primary>
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- $\lim _{x \rightarrow 0^{+}} x^x \quad\left(0^0\right.$ form)

- let $f(x)=x^x$. Then $\ln f(x)=x \ln x$

- We have seen that $\lim _{x \rightarrow 0^{+}} x \ln x=0$

- $\therefore \quad \lim _{x \rightarrow 0^{+}} \ln f(x)=0$.

- Now, $\quad f(x)=e^{\ln f(x)}$

- $ \therefore \lim _{x \rightarrow \infty} f(x)=\lim _{x \rightarrow \infty} e^{\ln f(x)} =e^{\text lim _{x \rightarrow 0^{+}} \ln f(x)} $

- $ =e^0=1 \text {. } \left(\because e^x\right.$ is continuous)

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<footer style = "font-size: 18px">$\infty-\infty $ form $\rightarrow$ $\frac{0}{0}$ form $\rightarrow$ <span style = "color:blue"> $ 0^0 $ form </span> $\rightarrow$ Remark $\rightarrow$ Example </footer>

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### <primary style="font-size:24px"> Remark </primary>
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- Remark: Note that if $\lim _{x \rightarrow c} \frac{f^{\prime}(x)}{g^{\prime}(x)}$ does not exist, we cannot conclude that $\lim _{x \rightarrow c} \frac{f(x)}{\partial(x)}$ does not exist.

- Example: $f(x)=x+\sin x ; g(x)=x$

- $\begin{aligned}& \lim _{x \rightarrow \infty} f(x)=\infty=\lim _{x \rightarrow \infty} g(x) \\\\& f^{\prime}(x)=1+\cos x, g^{\prime}(x)=1\end{aligned} $

- $\lim \frac{f^{\prime}(x)}{g^{\prime}(x)}=\lim _{x \rightarrow \infty}(1+\cos x)$, which does not exist.

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<footer style = "font-size: 18px">$\frac{0}{0}$ form $\rightarrow$ $ 0^0 $ form $\rightarrow$ <span style = "color:blue"> Remark </span> $\rightarrow$ Example $\rightarrow$ Thank you </footer>

### <Success style="font-size:25px"> Derivatives L-12 </Success>
### <primary style="font-size:24px"> Example </primary>
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- However,

- $\text { e. } \begin{aligned}& \lim _{x \rightarrow \infty} \frac{f(x)}{g(x)}=\lim _{x \rightarrow \infty} \frac{x+\sin x}{x} \\\\&=\lim _{x \rightarrow \infty}\left[1+\frac{\sin x}{x}\right] \\\\& 0 \leq\left|\frac{\sin x}{x}\right| \leq \frac{1}{x} \rightarrow 0 \text { is } x \rightarrow \infty .\end{aligned} $

- By the Sandwich the, $\lim _{x \rightarrow \infty} \frac{\sin x}{x}=0$

- $\therefore \lim _{x \rightarrow \infty} \frac{f(x)}{g(x)}=1+0=1 \text {. }$
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<footer style = "font-size: 18px">$ 0^0 $ form $\rightarrow$ Remark $\rightarrow$ <span style = "color:blue"> Example </span> $\rightarrow$ Thank you $\rightarrow$  </footer>