Evaluate the following limit: _ x 0 (1+x)^ 6 -1 (1+x)^2-1 - Vidyadip

Question

Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{(1+\text{x})^{{6}}-1}{(1+\text{x})^2-1}$

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Answer

$\lim\limits_{\text{x}\rightarrow0}\frac{(1+\text{x})^{{6}}-1}{(1+\text{x})^2-1}$$=\lim\limits_{\text{x}\rightarrow0}\frac{\frac{(1+\text{x})^{{{6}}}-1^6}{1+\text{x}-1}}{\frac{(1+\text{x})^2-1^2}{1+\text{x}-1}}$
$\Rightarrow \text{Let} 1 + \text{x} = \text{y}, \text{as x} → 0, \text{y} → 1$
$=\frac{\lim\limits_{\text{y}\rightarrow1}\frac{\text{y}^6-1^6}{\text{y}-1}}{\lim\limits_{\text{y}\rightarrow1}\frac{\text{y}^2-1}{\text{y}-1}}$
$=\frac{6(1)^{6-1}}{2(1)^{2-1}}$ $\Big[\text{Using formula} \lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^\text{n}-\text{a}^\text{n}}{\text{a}-\text{a}}=\text{na}^{\text{n}-1}\Big]$
$=\frac62$
$=3$

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