Question

Simplify the following

$log_{x \: \to \: 0}(tan( \frac{\pi}{4} + x) )^{ \frac{1}{x} }$
​

Answer 1

Appropriate Question :- Simplify the following:

$\rm \: \displaystyle\lim_{x \to 0}\rm {\bigg[tan\bigg(\dfrac{\pi}{4} + x\bigg) \bigg]}^{ \dfrac{1}{x} }$

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

Given expression is

$\rm \: \displaystyle\lim_{x \to 0}\rm {\bigg[tan\bigg(\dfrac{\pi}{4} + x\bigg) \bigg]}^{ \dfrac{1}{x} }$

If we substitute directly x = 0, we get

$\rm \: = {\bigg[tan\bigg(\dfrac{\pi}{4} + 0\bigg) \bigg]}^{ \dfrac{1}{0} }$

$\rm \: = {1}^{ \infty }$

which is indeterminant form.

So, Consider again

$\rm \: \displaystyle\lim_{x \to 0}\rm {\bigg[tan\bigg(\dfrac{\pi}{4} + x\bigg) \bigg]}^{ \dfrac{1}{x} }$

We know,

$\boxed{ \rm{ \:tan(x + y) = \frac{tanx + tany}{1 - tanx \: tany} \: }}$

So, using this result, we get

$\rm \: = \displaystyle\lim_{x \to 0}\rm {\bigg[\dfrac{tan \dfrac{\pi}{4} + tanx }{1 - tan \dfrac{\pi}{4} \: tanx} \bigg]}^{\dfrac{1}{x} }$

$\rm \: = \displaystyle\lim_{x \to 0}\rm {\bigg[\dfrac{1 + tanx }{1 - 1 \times \: tanx} \bigg]}^{\dfrac{1}{x} }$

$\rm \: = \displaystyle\lim_{x \to 0}\rm {\bigg[\dfrac{1 + tanx }{1 - tanx} \bigg]}^{\dfrac{1}{x} }$

$\rm \: = \displaystyle\lim_{x \to 0}\rm {\bigg[\dfrac{1 - tanx + tanx + tanx }{1 - tanx} \bigg]}^{\dfrac{1}{x} }$

$\rm \: = \displaystyle\lim_{x \to 0}\rm {\bigg[\dfrac{1 - tanx + 2tanx }{1 - tanx} \bigg]}^{\dfrac{1}{x} }$

$\rm \: = \displaystyle\lim_{x \to 0}\rm {\bigg[1 + \dfrac{ 2tanx }{1 - tanx} \bigg]}^{\dfrac{1}{x} }$

can be rewritten as

$\rm \: = \displaystyle\lim_{x \to 0}\rm {\bigg[1 + \dfrac{ 2tanx }{1 - tanx} \bigg]}^{\dfrac{1 - tanx}{2tanx} \times \dfrac{2tanx}{1 - tanx} \times \dfrac{1}{x} }$

We know,

$\boxed{ \rm{ \:\displaystyle\lim_{x \to 0}\rm {\bigg(1 + x \bigg) }^{\dfrac{1}{x} } = e \: }}$

So, using this result, we get

$\rm \: = \: {e}^{\displaystyle\lim_{x \to 0}\rm \frac{2tanx}{(1 - tanx)x} }$

$\rm \: = \: {e}^{\displaystyle\lim_{x \to 0}\rm \frac{2}{(1 - tanx)} \times\displaystyle\lim_{x \to 0}\rm \dfrac{tanx}{x} }$

$\rm \: = \: {e}^{2 \times 1}$

$\rm \: = \: {e}^{2}$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\rm \: \displaystyle\lim_{x \to 0}\rm {\bigg[tan\bigg(\dfrac{\pi}{4} + x\bigg) \bigg]}^{ \dfrac{1}{x} } = {e}^{2} \: }}$

$\rule{190pt}{2pt}$

Additional information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{sinx}{x} = 1}\\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{tanx}{x} = 1}\\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{log(1 + x)}{x} = 1}\\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{ {e}^{x} - 1}{x} = 1}\\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{ {a}^{x} - 1}{x} = loga}\\ \\ \end{array} }}\end{gathered}\end{gathered}\end{gathered}$