Question

lim x pi/ 2 (secx-tanx)/ (pi /2 -x)​

Answer 1

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

Given expression is

$\displaystyle\lim_{x \to \dfrac{\pi}{2}}\rm \: \frac{secx - tanx}{\dfrac{\pi}{2} - x}$

If we substitute directly the value of x, we get

$\rm \: = \: \dfrac{sec\dfrac{\pi}{2} - tan\dfrac{\pi}{2}}{\dfrac{\pi}{2} - \dfrac{\pi}{2}}$

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

which is indeterminant form.

So, Consider again

$\displaystyle\lim_{x \to \dfrac{\pi}{2}}\rm \: \frac{secx - tanx}{\dfrac{\pi}{2} - x}$

To evaluate this limit, we use Method of Substitution.

So, Substitute

$\rm \: x = \dfrac{\pi}{2} - h, \: \: as \: x \: \to \: \dfrac{\pi}{2}, \: \: so \: h \: \to \: 0$

So, on substituting this in above step, we get

$\rm \: = \: \displaystyle\lim_{h \to 0}\rm \frac{sec\bigg(\dfrac{\pi}{2} - h\bigg) - tan \bigg(\dfrac{\pi}{2} - h \bigg) }{\dfrac{\pi}{2} - \dfrac{\pi}{2} + h}$

$\rm \: = \: \displaystyle\lim_{h \to 0}\rm \frac{cosech - coth}{h}$

can be further rewritten as

$\rm \: = \: \displaystyle\lim_{h \to 0}\rm \frac{1}{h}\bigg[\dfrac{1}{sinh} - \frac{cosh}{sinh} \bigg]$

$\rm \: = \: \displaystyle\lim_{h \to 0}\rm \frac{1}{h}\bigg[\dfrac{1 - cosh}{sinh}\bigg]$

We know,

$\boxed{\tt{ 1 - cos2x = {2sin}^{2}x}} \\ \\ and \\ \\ \boxed{\tt{ sin2x = 2 \: sinx \: cosx \: }}$

So, using these Identities, we get

$\rm \: = \: \displaystyle\lim_{h \to 0}\rm \frac{2 {sin}^{2} \dfrac{h}{2} }{h \: \bigg(2 \: sin\dfrac{h}{2} \: cos\dfrac{h}{2}\bigg)}$

$\rm \: = \: \displaystyle\lim_{h \to 0}\rm \frac{tan\dfrac{h}{2}}{h}$

can be further rewritten as

$\rm \: = \: \displaystyle\lim_{h \to 0}\rm \frac{tan\dfrac{h}{2}}{\dfrac{h}{2} \times 2}$

$\rm \: = \: \dfrac{1}{2} \displaystyle\lim_{h \to 0}\rm \frac{tan\dfrac{h}{2}}{\dfrac{h}{2}}$

We know,

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

So, using this result, we get

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

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

Hence,

$\rm\implies \:\boxed{\tt{ \: \displaystyle\lim_{x \to \dfrac{\pi}{2}}\rm \: \frac{secx - tanx}{\dfrac{\pi}{2} - x} = \frac{1}{2} }}$

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ADDITIONAL INFORMATION

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

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

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

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

Answer 2

$\huge\green{Solution:-}$

• Please see the attached file