Question

tell both the questions​

Answer 1

$\displaystyle\mathbf{16.}\ \ \lim_{x\to \frac{\pi}{2}}\ (\sin x)^{\tan x}$

On applying the value of x directly, we get,

$\left(\sin\left(\dfrac{\pi}{2}\right)\right)^{\tan(\frac{\pi}{2})}\ =\ 1^{^{\frac{1}{0}}}\ =\ 1^{^{\infty}}$

So we have to get it in indeterminate form. We can!

We can write the given as,

$\displaystyle\lim_{x\to \frac{\pi}{2}}e^{\tan x\ln(\sin x)}$

We just applied  $a^b=e^{b\ln a}$

So,

$\displaystyle\normalsize\text{ \lim_{x\to \frac{\pi}{2}}e^{\tan x\cdot\ln(\sin x)}=\lim_{x\to \frac{\pi}{2}}\ e^{\frac{\ln(\sin x)}{\cot x}} }=e^{\displaystyle\scriptsize\text{ \lim_{x\to \frac{\pi}{2}}\ \frac{\ln(\sin x)}{\cot x} }}$

Since e is continuous, this is possible.

Now, take the limit only.

$\displaystyle\lim_{x\to \frac{\pi}{2}}\ \frac{\ln(\sin x)}{\cot x}$

Now it's in indeterminate form.

$\dfrac{\ln(\sin (\frac{\pi}{2}))}{\cot (\frac{\pi}{2})}=\dfrac{\ln(1)}{0}=\dfrac{0}{0}$

So we can apply L'Hospital's Rule.

$\displaystyle\lim_{x\to \frac{\pi}{2}}\ \frac{\ln(\sin x)}{\cot x}\ =\ \lim_{x\to\frac{\pi}{2}}\ \dfrac{\frac{\cos x}{\sin x}}{-\csc^2x}\ =\ \lim_{x\to\frac{\pi}{2}}\ -\dfrac{\cot x}{\csc^2x}\\ \\ \\ =\ \lim_{x\to\frac{\pi}{2}}\ -\dfrac{\sin^2x}{\tan x}\ =\ \lim_{x\to\frac{\pi}{2}}\ -\sin x\cdot\cos x\\ \\ \\ =\ -\sin\left(\frac{\pi}{2}\right)\cdot\cos\left(\frac{\pi}{2}\right)\ =\ -1\cdot 0=0$

So we got the limit 0. Now,

$\displaystyle e^{\displaystyle\scriptsize\text{ \lim_{x\to \frac{\pi}{2}}\ \frac{\ln(\sin x)}{\cot x} }}\ =\ e^0\ =\ \mathbf{1}$

Hence 1 is the answer.

$\mathbf{18.}\ \ \displaystyle\lim_{x\to0}\ (1+\sin x)^{\cot x}$

Here we also do what we did earlier!

$\displaystyle\lim_{x\to0}\ (1+\sin x)^{\cot x}\ =\ \lim_{x\to0}\ e^{\cot x\ln(1+\sin x)}\\ \\ \\ =\ \lim_{x\to0}\ e^{\frac{\ln(1+\sin x)}{\tan x}}\ =\ e^{\displaystyle\scriptsize\lim_{x\to0}\ \frac{\ln(1+\sin x)}{\tan x}}$

Taking the limit, applying L'Hospital's Rule,

$\displaystyle\lim_{x\to0}\ \frac{\ln(1+\sin x)}{\tan x}\ =\ \lim_{x\to0}\ \dfrac{\frac{\cos x}{1+\sin x}}{\sec^2x}\ =\ \lim_{x\to0}\ \frac{\cos^3x}{1+\sin x}\\ \\ \\ =\ \lim_{x\to0}\ \frac{(1-\sin^2x)\cos x}{\frac{1-\sin^2x}{1-\sin x}}\ =\ \lim_{x\to0}\ \frac{\cos x}{\frac{1}{1-\sin x}}\ =\ \lim_{x\to0}\ \cos x(1-\sin x)\\ \\ \\ =\ \cos0\ (1-\sin 0)\ =\ 1(1-0)\ =\ 1\cdot 1=1$

Now,

$e^{\displaystyle\scriptsize\lim_{x\to0}\ \frac{\ln(1+\sin x)}{\tan x}}\ =\ e^1\ =\ \mathbf{e}$

Hence e is the answer.

Answer 2

Answer:

$</p><p>\displaystyle\mathbf{16.}\ \ \lim_{x\to \frac{\pi}{2}}\ (\sin x)^{\tan x}16. </p><p>x→ </p><p>2</p><p>π</p><p>$

lim

(sinx)

tanx

$</p><p></p><p>On applying the value of x directly, we get,</p><p></p><p>(\sin(\dfrac{\pi}{2}))^{\tan(\frac{\pi}{2})}\ =\ 1^{^{\frac{1}{0}}}\ =\ 1^{^{\infty}}(sin( </p><p>2</p><p>π</p><p> </p><p> )) </p><p>tan( </p><p>2</p><p>π</p><p> </p><p> )</p><p> = 1 </p><p>0</p><p>1</p><p> </p><p> </p><p> </p><p> = 1 </p><p>∞</p><p> </p><p>$

$</p><p>So we have to get it in indeterminate form. We can!</p><p></p><p>We can write the given as,</p><p></p><p>\displaystyle\lim_{x\to \frac{\pi}{2}}e^{\tan x\ln(\sin x)} </p><p>x→ </p><p>2</p><p>π</p><p> </p><p> </p><p>lim</p><p> </p><p> e </p><p>tanxln(sinx)</p><p> </p><p></p><p>We just applied a^b=e^{b\ln a}a </p><p>b</p><p> =e </p><p>blna</p><p> </p><p></p><p>So,</p><p></p><p>\displaystyle\normalsize\text{ \lim_{x\to \frac{\pi}{2}}e^{\tan x\cdot\ln(\sin x)}=\lim_{x\to \frac{\pi}{2}}\ e^{\frac{\ln(\sin x)}{\cot x}} }=e^{\displaystyle\scriptsize\text{ \lim_{x\to \frac{\pi}{2}}\ \frac{\ln(\sin x)}{\cot x} }}lim </p><p>x→ </p><p>2</p><p>π</p><p> </p><p> </p><p> </p><p> e </p><p>tanx⋅ln(sinx)</p><p> =lim </p><p>x→ </p><p>2</p><p>π</p><p> </p><p> </p><p> </p><p> e </p><p>cotx</p><p>ln(sinx)</p><p> </p><p> </p><p> =e </p><p>lim </p><p>x→ </p><p>2</p><p>π</p><p> </p><p> </p><p> </p><p> </p><p>cotx</p><p>ln(sinx)</p><p> </p><p> </p><p>$

$</p><p>Since e is continuous, this is possible.</p><p></p><p>Now, take the limit only.</p><p></p><p>\displaystyle\lim_{x\to \frac{\pi}{2}}\ \frac{\ln(\sin x)}{\cot x} </p><p>x→ </p><p>2</p><p>π</p><p> </p><p> </p><p>lim</p><p> </p><p> </p><p>cotx</p><p>ln(sinx)</p><p> </p><p> </p><p></p><p>Now it's in indeterminate form.</p><p></p><p>\dfrac{\ln(\sin (\frac{\pi}{2}))}{\cot (\frac{\pi}{2})}=\dfrac{\ln(1)}{0}=\dfrac{0}{0} </p><p>cot( </p><p>2</p><p>π</p><p> </p><p> )</p><p>ln(sin( </p><p>2</p><p>π</p><p> </p><p> ))</p><p> </p><p> = </p><p>0</p><p>ln(1)</p><p> </p><p> = </p><p>0</p><p>0</p><p> </p><p> </p><p></p><p>So we can apply L'Hospital's Rule.</p><p></p><p>\begin{gathered}\displaystyle\lim_{x\to \frac{\pi}{2}}\ \frac{\ln(\sin x)}{\cot x}\ =\ \lim_{x\to\frac{\pi}{2}}\ \dfrac{\frac{\cos x}{\sin x}}{-\csc^2x}\ =\ \lim_{x\to\frac{\pi}{2}}\ -\dfrac{\cot x}{\csc^2x}\\ \\ \\ =\ \lim_{x\to\frac{\pi}{2}}\ -\dfrac{\sin^2x}{\tan x}\ =\ \lim_{x\to\frac{\pi}{2}}\ -\sin x\cdot\cos x\\ \\ \\ =\ -\sin(\frac{\pi}{2})\cdot\cos(\frac{\pi}{2})\ =\ -1\cdot 0=0\end{gathered} </p><p>x→ </p><p>2</p><p>π</p><p> </p><p> </p><p>lim</p><p> </p><p> </p><p>cotx</p><p>ln(sinx)</p><p> </p><p> = </p><p>x→ </p><p>2</p><p>π</p><p> </p><p> </p><p>lim</p><p> </p><p> </p><p>−csc </p><p>2</p><p> x</p><p>sinx</p><p>cosx</p><p> </p><p> </p><p> </p><p> = </p><p>x→ </p><p>2</p><p>π</p><p> </p><p> </p><p>lim</p><p> </p><p> − </p><p>csc </p><p>2</p><p> x</p><p>cotx</p><p> </p><p> </p><p>= </p><p>x→ </p><p>2</p><p>π</p><p> </p><p> </p><p>lim</p><p> </p><p> − </p><p>tanx</p><p>sin </p><p>2</p><p> x</p><p> </p><p> = </p><p>x→ </p><p>2</p><p>π</p><p> </p><p> </p><p>lim</p><p> </p><p> −sinx⋅cosx</p><p>= −sin( </p><p>2</p><p>π</p><p> </p><p> )⋅cos( </p><p>2</p><p>π</p><p> </p><p> ) = −1⋅0=0</p><p> </p><p> </p><p>$

$</p><p>So we got the limit 0. Now,</p><p></p><p>\displaystyle e^{\displaystyle\scriptsize\text{ \lim_{x\to \frac{\pi}{2}}\ \frac{\ln(\sin x)}{\cot x} }}\ =\ e^0\ =\ \mathbf{1}e </p><p>lim </p><p>x→ </p><p>2</p><p>π</p><p> </p><p> </p><p> </p><p> </p><p>cotx</p><p>ln(sinx)</p><p> </p><p> </p><p> = e </p><p>0</p><p> = 1</p><p></p><p>Hence 1 is the answer.</p><p></p><p>\mathbf{18.}\ \ \displaystyle\lim_{x\to0}\ (1+\sin x)^{\cot x}18. </p><p>x→0</p><p>lim</p><p> </p><p> (1+sinx) </p><p>cotx</p><p> </p><p></p><p>Here we also do what we did earlier!</p><p></p><p>\begin{gathered}\displaystyle\lim_{x\to0}\ (1+\sin x)^{\cot x}\ =\ \lim_{x\to0}\ e^{\cot x\ln(1+\sin x)}\\ \\ \\ =\ \lim_{x\to0}\ e^{\frac{\ln(1+\sin x)}{\tan x}}\ =\ e^{\displaystyle\scriptsize\lim_{x\to0}\ \frac{\ln(1+\sin x)}{\tan x}}\end{gathered} </p><p>x→0</p><p>lim</p><p> </p><p> (1+sinx) </p><p>cotx</p><p> = </p><p>x→0</p><p>lim</p><p> </p><p> e </p><p>cotxln(1+sinx)</p><p> </p><p>= </p><p>x→0</p><p>lim</p><p> </p><p> e </p><p>tanx</p><p>ln(1+sinx)</p><p> </p><p> </p><p> = e </p><p>lim </p><p>x→0</p><p> </p><p> </p><p>tanx</p><p>ln(1+sinx)</p><p> </p><p> </p><p> </p><p> </p><p> </p><p>$

$</p><p>Taking the limit, applying L'Hospital's Rule,</p><p></p><p>\begin{gathered}\displaystyle\lim_{x\to0}\ \frac{\ln(1+\sin x)}{\tan x}\ =\ \lim_{x\to0}\ \dfrac{\frac{\cos x}{1+\sin x}}{\sec^2x}\ =\ \lim_{x\to0}\ \frac{\cos^3x}{1+\sin x}\\ \\ \\ =\ \lim_{x\to0}\ \frac{(1-\sin^2x)\cos x}{\frac{1-\sin^2x}{1-\sin x}}\ =\ \lim_{x\to0}\ \frac{\cos x}{\frac{1}{1-\sin x}}\ =\ \lim_{x\to0}\ \cos x(1-\sin x)\\ \\ \\ =\ \cos0\ (1-\sin 0)\ =\ 1(1-0)\ =\ 1\cdot 1=1\end{gathered} </p><p>x→0</p><p>lim</p><p> </p><p> </p><p>tanx</p><p>ln(1+sinx)</p><p> </p><p> = </p><p>x→0</p><p>lim</p><p> </p><p> </p><p>sec </p><p>2</p><p> x</p><p>1+sinx</p><p>cosx</p><p> </p><p> </p><p> </p><p> = </p><p>x→0</p><p>lim</p><p> </p><p> </p><p>1+sinx</p><p>cos </p><p>3</p><p> x</p><p> </p><p> </p><p>= </p><p>x→0</p><p>lim</p><p> </p><p> </p><p>1−sinx</p><p>1−sin </p><p>2</p><p> x</p><p> </p><p> </p><p>(1−sin </p><p>2</p><p> x)cosx</p><p> </p><p> = </p><p>x→0</p><p>lim</p><p> </p><p> </p><p>1−sinx</p><p>1</p><p> </p><p> </p><p>cosx</p><p> </p><p> = </p><p>x→0</p><p>lim</p><p> </p><p> cosx(1−sinx)</p><p>= cos0 (1−sin0) = 1(1−0) = 1⋅1=1</p><p> </p><p> </p><p></p><p>Now,</p><p></p><p>e^{\displaystyle\scriptsize\lim_{x\to0}\ \frac{\ln(1+\sin x)}{\tan x}}\ =\ e^1\ =\ \mathbf{e}e </p><p>lim </p><p>x→0</p><p> </p><p> </p><p>tanx</p><p>ln(1+sinx)</p><p> </p><p> </p><p> = e </p><p>1</p><p> = e</p><p></p><p>Hence e is the answer.</p><p>$