Question

limit x extends to k Find (Tan x - Tan k) /(Sin x - Sin k)​

Answer 1

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

Given expression is

$\rm :\longmapsto\:\displaystyle\lim_{x \to k}\sf \frac{tanx - tank}{sinx - sink}$

If we put directly x = k, we get indeterminant form.

So, to evaluate this limit,

We know,

$\red{ \boxed{ \sf{ \:sinx - siny = 2cos\bigg[\dfrac{x + y}{2} \bigg]sin\bigg[\dfrac{x - y}{2} \bigg]}}}$

So, using this, we get,

$\rm \:  =  \: \displaystyle\lim_{x \to k}\sf \frac{ \dfrac{sinx}{cosx} - \dfrac{sink}{cosk} }{2cos\bigg[\dfrac{x + k}{2} \bigg]sin\bigg[\dfrac{x - k}{2} \bigg]}$

$\rm \:  =  \: \dfrac{1}{2cos\bigg[\dfrac{k + k}{2} \bigg]}\displaystyle\lim_{x \to k}\sf \frac{sinxcosk - sinkcosx}{cosk \: cosx \times sin\bigg[\dfrac{x - k}{2} \bigg]}$

We know,

$\red{ \boxed{ \sf{ \:sinx \: cosy - siny \: cosx = sin(x - y)}}}$

So, using this identity, we get

$\rm \:  =  \: \dfrac{1}{2cos\bigg[\dfrac{2k}{2} \bigg] \times cosk \: cosk}\displaystyle\lim_{x \to k}\sf \frac{sin(x - k)}{sin\bigg[\dfrac{x - k}{2} \bigg]}$

We know,

$\red{ \boxed{ \sf{ \:sin2x = 2sinx \: cosx}}}$

So, using this identity, we get

$\rm \:  =  \: \dfrac{1}{2 {cos}^{3}k} \: \displaystyle\lim_{x \to k}\sf \frac{2sin\bigg[\dfrac{x - k}{2} \bigg]cos\bigg[\dfrac{x - k}{2} \bigg]}{sin\bigg[\dfrac{x - k}{2} \bigg]}$

$\rm \:  =  \: \dfrac{1}{{cos}^{3}k} \: \displaystyle\lim_{x \to k}\sf cos\bigg[\dfrac{x - k}{2} \bigg]$

$\rm \:  =  \: \dfrac{1}{{cos}^{3}k} \: \times \sf cos\bigg[\dfrac{k - k}{2} \bigg]$

$\rm \:  =  \: \dfrac{1}{{cos}^{3}k} \: \times \sf cos\bigg[\dfrac{0}{2} \bigg]$

$\rm \:  =  \: \dfrac{1}{{cos}^{3}k} \: \times \sf 1$

$\rm \:  =  \: \dfrac{1}{{cos}^{3}k}$

$\rm \:  =  \: {sec}^{3}k$

Hence,

$\rm :\longmapsto\:\red{ \boxed{ \sf{ \: \: \:\displaystyle\lim_{x \to k}\sf \frac{tanx - tank}{sinx - sink} = {sec}^{3}k \: \: \: }}}$

Additional Information :-

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

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

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

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

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