Question

Evaluate the limit. ​

Answer 1

Given Question

Evaluate the limit

$\displaystyle\lim_{x \to 0} \frac{ {\bigg(1 + x\bigg)}^{ \dfrac{1}{m} } - {\bigg(1 + x\bigg)}^{ \dfrac{1}{n} } }{x}$

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

Consider,

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0} \frac{ {\bigg(1 + x\bigg)}^{ \dfrac{1}{m} } - {\bigg(1 + x\bigg)}^{ \dfrac{1}{n} } }{x}$

If we substitute directly x = 0, we get

$\rm \: = \:\ \dfrac{ {\bigg(1 + 0\bigg)}^{ \dfrac{1}{m} } - {\bigg(1 + 0\bigg)}^{ \dfrac{1}{n} } }{0}$

$\rm \: = \:\ \dfrac{ {\bigg(1\bigg)}^{ \dfrac{1}{m} } - {\bigg(1\bigg)}^{ \dfrac{1}{n} } }{0}$

$\rm \: = \:\dfrac{1 - 1}{0}$

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

which is indeterminant form.

So, To evaluate, we use method of expansions.

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0} \frac{ {\bigg(1 + x\bigg)}^{ \dfrac{1}{m} } - {\bigg(1 + x\bigg)}^{ \dfrac{1}{n} } }{x}$

We know,

From binomial expansion, we have

$\boxed{ \tt{ \: {(1 + x)}^{n} = 1 + nx + \frac{n(n - 1)}{2} {x}^{2} + - - - \: }}$

So, using this, identity, we get

$\rm=\displaystyle\lim_{x \to 0}\dfrac{\bigg[1 +\dfrac{1}{m}x + \dfrac{m(m - 1)}{2} {x}^{2} + - - - \bigg] - \bigg[1+\dfrac{1}{n}x+\dfrac{n(n - 1)}{2} {x}^{2} + - - - \bigg]}{x}$

$\rm=\displaystyle\lim_{x \to 0}\dfrac{\bigg[\dfrac{1}{m}x + \dfrac{m(m - 1)}{2} {x}^{2} + - - - \bigg] - \bigg[\dfrac{1}{n}x+\dfrac{n(n - 1)}{2} {x}^{2} + - - - \bigg]}{x}$

$\rm=\displaystyle\lim_{x \to 0}\dfrac{x\bigg \{\bigg[\dfrac{1}{m} + \dfrac{m(m - 1)}{2} {x}^{} + - - - \bigg] - \bigg[\dfrac{1}{n}+\dfrac{n(n - 1)}{2} {x}^{} + - - - \bigg]\bigg \}}{x}$

$\rm=\displaystyle\lim_{x \to 0}\bigg \{\bigg[\dfrac{1}{m} + \dfrac{m(m - 1)}{2} {x}^{} + - - - \bigg] - \bigg[\dfrac{1}{n}+\dfrac{n(n - 1)}{2} {x}^{} + - - - \bigg]\bigg \}$

$\rm \: = \:\dfrac{1}{m} - \dfrac{1}{n}$

$\rm \: = \:\dfrac{n - m}{mn}$

Hence,

$\boxed{ \tt{ \: \displaystyle\lim_{x \to 0} \frac{ {\bigg(1 + x\bigg)}^{ \dfrac{1}{m} } - {\bigg(1 + x\bigg)}^{ \dfrac{1}{n} } }{x} = \frac{n - m}{nm} \: }}$

More to know :-

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

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