Question

Lim. 2x-π\cosx
X=1\2 ​

Answer 1

Question:

$\text{Find\ \ \ \displaystyle\lim_{x\to \frac{\pi}{2}}\ \dfrac{2x-\pi}{\cos(x)} .}$

Solution:

On finding the limit directly by taking  $x=\dfrac{\pi}{2},$  we get it as indeterminate form.

$\dfrac{2\cdot\frac{\pi}{2}-\pi}{\cos(\frac{\pi}{2})}\ =\ \dfrac{\pi-\pi}{\cos(\frac{\pi}{2})}\ =\ \dfrac{0}{0}$

So we have to use L'hospital's Rule.

$\displaystyle\lim_{x\to a}\ \dfrac{f(x)}{g(x)}\ =\ \lim_{x\to a}\ \dfrac{f'(x)}{g'(x)}\ \ \iff\ \ \dfrac{f(x)}{g(x)}=\dfrac{0}{0}\ \ \text{OR}\ \ \dfrac{f(x)}{g(x)}=\dfrac{\pm\infty}{\pm\infty}$

So we have to derive both the numerator and denominator of the fraction whose limit is given to find.

Consider the numerator.

$\dfrac{d}{dx}\ (2x-\pi)=2\cdot \dfrac{d}{dx}\ x-\ \dfrac{d}{dx}\ \pi\ =\ 2\cdot 1-0\ =\ 2$

Then consider the denominator.

$\dfrac{d}{dx}(\cos(x))=-\sin(x)$

Now,

$\displaystyle\lim_{x\to \frac{\pi}{2}}\ \frac{2x-\pi}{\cos(x)}\ =\ \lim_{x\to \frac{\pi}{2}}\ \frac{2}{-\sin(x)}\ =\ \dfrac{2}{-\sin(\frac{\pi}{2})}\ =\ \dfrac{2}{-1}=\bold{-2}$

Hence -2 is the answer.

Answer 2

$\huge\underline\textbf{Answer:-}$

Refer the given attachment:-