Question

lim x tend to pi/2 cosx/log(x-pi/2+1)​

Answer 1

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm :\longmapsto\:\displaystyle\lim_{x \to \dfrac{\pi}{2} } \: \frac{cosx}{log\bigg[x - \dfrac{\pi}{2} + 1\bigg]}$

On substituting directly the value of x, we get

$\rm \:  =  \: \: \dfrac{cos\dfrac{\pi}{2} }{log\bigg[\dfrac{\pi}{2} - \dfrac{\pi}{2} + 1\bigg]}$

$\rm \:  =  \: \dfrac{0}{log1}$

$\rm \:  =  \: \dfrac{0}{0}$

which is indeterminant form.

Consider, again

$\rm :\longmapsto\:\displaystyle\lim_{x \to \dfrac{\pi}{2} } \: \frac{cosx}{log\bigg[x - \dfrac{\pi}{2} + 1\bigg]}$

To evaluate this limit, we use Method of Substitution.

So, Substitute

$\purple{\rm :\longmapsto\:x = \dfrac{\pi}{2} + y}$

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

So, above expression can be rewritten as

$\rm \:  =  \: \displaystyle\lim_{y \to 0 } \: \frac{cos\bigg[\dfrac{\pi}{2} + y\bigg]}{log\bigg[ \dfrac{\pi}{2} + y - \dfrac{\pi}{2} + 1\bigg]}$

$\rm \:  =  \: \displaystyle\lim_{y \to 0} \frac{ - \: siny}{log(1 + y)}$

$\rm \:  =  \: - \: \displaystyle\lim_{y \to 0} \frac{\: siny}{log(1 + y)}$

$\rm \:  =  \: - \: \displaystyle\lim_{y \to 0} \: \frac{siny}{y} \: \times \: \displaystyle\lim_{y \to 0} \frac{y}{log(1 + y)}$

$\rm \:  =  \: - 1 \times 1$

$\rm \:  =  \: - 1$

Hence,

$\rm :\longmapsto\:\boxed{\tt{ \: \displaystyle\lim_{x \to \dfrac{\pi}{2} } \: \frac{cosx}{log\bigg[x - \dfrac{\pi}{2} + 1\bigg]} = - 1}}$

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Formula Used

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

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

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More to know

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

$\purple{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x\to 0} \frac{ {a}^{x} - 1}{x} = loga}}}$

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