Question

${ \underline{ \frak{ \red{Question}}}}$

Evaluate :
$\ \sf{\rm \displaystyle \lim_{x \to 0}\ \sf{\bigg(1-\frac{3x}{5}\bigg)^{\frac{1}{x}}}}$

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Answer 1

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

Given expression is

$\rm \: \sf{\rm \displaystyle \lim_{x \to 0}\ \sf{\bigg(1-\frac{3x}{5}\bigg)^{\dfrac{1}{x}}}}$

can be rewritten as

$\rm \: = \: \sf{\rm \displaystyle \lim_{x \to 0}\ \sf{\bigg(1 + \frac{ - 3x}{5}\bigg)^{\dfrac{1}{x}}}}$

can be further rewritten as

$\rm \: = \: \sf{\rm \displaystyle \lim_{x \to 0}\ \sf{\bigg(1 + \frac{ - 3x}{5}\bigg)^{\dfrac{5}{ - 3x} \times \dfrac{ - 3x}{5} \times \dfrac{1}{x}}}}$

can be further rewritten as

$\rm \: = \: \sf{\rm \displaystyle \lim_{x \to 0}\ \sf{\bigg(1 + \frac{ - 3x}{5}\bigg)^{\dfrac{5}{ - 3x} \times \dfrac{ - 3}{5}}}}$

We know,

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

So, using this result, we get

$\rm \: =  \: {\bigg(e\bigg) }^{\dfrac{ - 3}{5} }$

Hence,

$\rm\implies \:\rm \: \boxed{\sf{  \:\sf{\rm \displaystyle \lim_{x \to 0}\ \sf{\bigg(1-\frac{3x}{5}\bigg)^{\dfrac{1}{x}}}} \: =  \: {\bigg(e\bigg) }^{\dfrac{ - 3}{5} } \: \: }}$

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Additional Information :-

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

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

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

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

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

Answer 2

$\large\underline{\sf{Solution-}} \\ \\ \begin{gathered}\rm \: \sf{\rm \displaystyle \lim_{x \to 0}\ \sf{\bigg(1-\frac{3x}{5}\bigg)^{\dfrac{1}{x}}}} \\ \end{gathered} \\ \\ can \: \: be \: \: rewritten \: \: as \\ \\\begin{gathered}\rm \: = \: \sf{\rm \displaystyle \lim_{x \to 0}\ \sf{\bigg(1 + \frac{ - 3x}{5}\bigg)^{\dfrac{1}{x}}}} \\ \end{gathered} \\ \\ can \: \: be \: \: further \: \: rewritten \: \: as \\ \\ \begin{gathered}\rm \: = \: \sf{\rm \displaystyle \lim_{x \to 0}\ \sf{\bigg(1 + \frac{ - 3x}{5}\bigg)^{\dfrac{5}{ - 3x} \times \dfrac{ - 3x}{5} \times \dfrac{1}{x}}}} \\ \end{gathered} \\ \\ can \: \: be \: \: further \: \: rewritten \: \: as \\ \\ \begin{gathered}\rm \: = \: \sf{\rm \displaystyle \lim_{x \to 0}\ \sf{\bigg(1 + \frac{ - 3x}{5}\bigg)^{\dfrac{5}{ - 3x} \times \dfrac{ - 3}{5}}}} \\\end{gathered} \\ \\ We \: \: know, \\ \\\begin{gathered}\boxed{\sf{ \:\displaystyle \lim_{x \to 0}\rm \: {\bigg(1 + x \bigg) }^{\dfrac{1}{x} } = e \: \: }} \\ \end{gathered} \\\\So \: \: , using \: \: this \: \: result \: \: , \: \: we \: \: get \\ \\\begin{gathered}\rm \: = \: {\bigg(e\bigg) }^{\dfrac{ - 3}{5} } \\ \end{gathered} \\ \\ Hence, \: \:\\\\ \begin{gathered}\rm\implies \:\rm \: \boxed{\sf{ \:\sf{\rm \displaystyle \lim_{x \to 0}\ \sf{\bigg(1-\frac{3x}{5}\bigg)^{\dfrac{1}{x}}}} \: = \: {\bigg(e\bigg) }^{\dfrac{ - 3}{5} } \: \: }} \\ \end{gathered}$