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Home  >>  CBSE XI  >>  Math  >>  Limits and Derivatives
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Evaluate the following limits \[\] $ \lim\limits_{x \to 0} \large\frac{(x+1)^5-1}{x}$

$\begin{array}{1 1}5 \\ \large\frac{4}{5} \\ \large\frac{5}{4} \\ 4 \end{array} $

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$ \lim\limits_{x \to 0} \large\frac{(x+1)^5-1}{x}$
If we substitute $ x + 1 = y$, then $y \rightarrow 1 $ becomes $ x \rightarrow 0$.
$\Rightarrow \lim\limits_{x \to 0} \large\frac{(x+1)^5-1}{x}$ $=\lim\limits_{y \to 1} \large\frac{y^5-1}{y-1}$
$\qquad = \lim\limits_{y \to 1} \large\frac{y^5-1^5}{y-1}$
$\qquad = 5.1^{5-1} \quad \quad \bigg [ \lim\limits_{ x \to a } \large\frac{x^n-a^n}{x-a} = na^{n-1} \bigg]$
$\qquad = 5$
answered Apr 3, 2014 by thanvigandhi_1
edited May 16, 2014 by balaji.thirumalai
 

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