Evaluate the following limits: _ x 0 [(1-x)^8-1 (1-x)^2-1 ] - Vidyadip

Question

Evaluate the following limits: $\lim _{x \rightarrow 0}\left[\frac{(1-x)^8-1}{(1-x)^2-1}\right]$

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Answer

Put $1-x=y$
As $x \rightarrow 0, y \rightarrow 1$
$\therefore \lim _{x \rightarrow 0} \frac{(1-x)^8-1}{(1-x)^2-1}$
$ =\lim _{y \rightarrow 1} \frac{y^8-1^8}{y^2-1^2}$
$=\lim _{y \rightarrow 1} \frac{\frac{y^8-1^8}{y-1}}{\frac{y^2-1^2}{y-1}} \\ \cdots\left[\begin{array}{l} \because y \rightarrow 1, y \neq 1 \\ \therefore y-1 \neq 0 \end{array}\right]$
$=\frac{\lim _{y \rightarrow 1} \frac{y^8-1^8}{y-1}}{\lim _{y \rightarrow 1} \frac{y^2-1^2}{y-1}}$
$=\frac{8(1)^7}{2(1)^1} \quad \cdots\left[\because \lim _{x \rightarrow a} \frac{x^n-a^n}{x-a}=n . a^{n-1}\right]$
$=4 $

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