Question

how to solve it ??.step by step please ​

Answer 1

$\large\underline{\sf{Given \:Question - }}$

Evaluate the following

$\rm :\longmapsto\:\displaystyle\lim_{x \to \dfrac{\pi}{2} } \: \frac{cosx}{\pi - 2x}$

$\purple{ \bf{ \: \large\underline{\sf{Solution-}}}}$

Given expression is

$\rm :\longmapsto\:\displaystyle\lim_{x \to \dfrac{\pi}{2} } \: \frac{cosx}{\pi - 2x}$

On directly Substitute the limit, we get

$\rm \: = \:\dfrac{cos\dfrac{\pi}{2}}{\pi - 2 \times \dfrac{\pi}{2}}$

$\rm \: = \:\dfrac{0}{0}$

$\rm \: = \:which \: is \: meaningless.$

So, evaluate these, type of

$\rm :\longmapsto\:\displaystyle\lim_{x \to \dfrac{\pi}{2} } \: \frac{cosx}{\pi - 2x}$

We use Method of Substitution

$\purple{ \rm :\longmapsto\:\bf{ \: Put \: x \: = \: \dfrac{\pi}{2} - y \: as \: x \: to \: \dfrac{\pi}{2}, \: so \: y \: \to \: 0 }}$

So, above expression can be rewritten as

$\rm \: = \:\displaystyle\lim_{y \to 0 } \: \dfrac{cos\bigg[\dfrac{\pi}{2} - y\bigg]}{\pi - 2 \times \bigg[\dfrac{\pi}{2} - y \bigg]}$

$\rm \: = \:\displaystyle\lim_{y \to 0 } \: \dfrac{siny}{\pi - \pi + 2y}$

$\rm \: = \:\displaystyle\lim_{y \to 0 } \: \dfrac{siny}{ 2y}$

$\rm \: = \:\dfrac{1}{2} \displaystyle\lim_{y \to 0 } \: \dfrac{siny}{ y}$

$\rm \: = \:\dfrac{1}{2}$

Hence,

$\rm :\longmapsto\ \red{\boxed{ \rm{ \: \: \: \displaystyle\lim_{x \to \dfrac{\pi}{2} } \: \frac{cosx}{\pi - 2x} = \frac{1}{2} \: \: \: }}}$

More to.know

$\red{\boxed{ \rm{\displaystyle\lim_{x \to 0 } \: \: \frac{sinx}{x} = 1 \: \: }}}$

$\red{\boxed{ \rm{\displaystyle\lim_{x \to 0 } \: \: \frac{sin {}^{ - 1} x}{x} = 1 \: \: }}}$

$\red{\boxed{ \rm{\displaystyle\lim_{x \to 0 } \: \: \frac{ {e}^{x} - 1}{x} \: \: = 1}}}$

$\red{\boxed{ \rm{\displaystyle\lim_{x \to 0 } \: \: \frac{ log(1 + x)}{x} \: \: = 1}}}$

$\red{\boxed{ \rm{\displaystyle\lim_{x \to 0 } \: \: \frac{tanx}{x} \: \: = 1}}}$

Answer 2

Note: Here I will solve this problem by L'Hospitals rules.So that you can get two types of method of the solution.[ One type of method is given by 'mathdude500']

Solution:

L'Hospitals rules:

L’Hospital’s Rule tells us that if we have an indeterminate form 0/0 or ∞/∞ all we need to do is differentiate the numerator and differentiate the denominator until we get determinate form and then take the limit.

According to the question,

In the limit if we plugged in x=π/2, we would get 0/0.

Cause :

$\blue{ \boxed{\boxed{\begin{array}{cc}\rm \displaystyle\lim_{ x \to \frac{\pi}{2}} \rm \frac{cos \: x}{\pi - 2x} \\ \\ = \frac{cos \frac{\pi}{2} }{\pi - 2 \times \frac{\pi}{2} } \\ \\ = \frac{0}{\pi - \pi} \\ \\ = \frac{0}{0} \end{array}}}}</p><p>$

It is called indeterminate forms.Here we can apply that rules.

Now,

$\orange{ \boxed{\boxed{\begin{array}{cc}\rm \to \:given \\ \\ \rm \displaystyle\lim_{ x \to \frac{\pi}{2}} \rm \: \frac{cos \: x}{\pi - 2x} \\ \\ \red{ \sf \: apply \: L Hospital \: Rule }\ \\ \\ = \rm \displaystyle\lim_{ x \to \frac{\pi}{2}} \rm \frac{ \frac{d }{dx} cos \: x}{ \frac{d}{dx} (\pi - 2x)} \\ \\ \underbrace{\pink{ {\boxed{\begin{array}{cc}\sf \to \:we \: know \: that \\ \\ \sf \hookrightarrow \: \frac{d}{dx} cos \: x = - sin \: x \\ \\ \sf \hookrightarrow \: \frac{d}{dx}(constant) = 0 \\ \\ \sf \hookrightarrow \: \frac{d}{dx} {x}^{n} = n. {x}^{n - 1} \\ \\ \sf \hookrightarrow \: \frac{d}{dx} (u \pm \: v) = \frac{d \: u}{dx} \pm \frac{dv}{dx} \end{array}}}}}_ {\red \: \sf\: apply \: these \: formula \: \downarrow} \\ \\ = \rm \displaystyle\lim_{ x \to \frac{\pi}{2}} \rm \frac{ - sin \: x}{ \frac{d \: \pi}{dx} - \frac{d}{dx} 2x } \\ \\ = \rm \displaystyle\lim_{ x \to \frac{\pi}{2}} \rm \frac{ - sin \: x}{0 - 2 \times 1} \\ \\ = \rm \displaystyle\lim_{ x \to \frac{\pi}{2}} \rm \frac{ sin \: x}{ 2} \\ \\ \red{ \sf \: apply \: limit} \\ \\ = \rm \: \frac{sin \frac{\pi}{2} }{2} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{1}{2} \: \: \{ \because \: sin \frac{\pi}{2} = 1 \} \\ \\ \\ \blue{ \boxed{\therefore \: \rm \displaystyle\lim_{ x \to \frac{\pi}{2}} \rm \frac{cos \: x}{\pi - 2x} = \frac{1}{2}}} \end{array}}}}</p><p>$