Question

lim x tend to 0 (1+x)^5-1/(1+x)^3-1​

Answer 1

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used \::}}}} \end{gathered}$

$\large \boxed{\tt \:  ⟼ \: \lim_{x\to a}\dfrac{ {x}^{n} - {a}^{n} }{x - a} = n {a}^{n - 1} }$

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$\large\underline\blue{\bold{Solution :- }}$

$\tt \:  ⟼ \: \lim_{x\to 0}\dfrac{ {(1 + x)}^{5} - 1}{ {(1 + x)}^{3} - 1 }$

$\large\underline{\bold{❥︎Step :- 1 }}$

☆ If we substitute x = 0, in the given statement, we get indeterminant form 0/0.

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$\large\underline{\bold{❥︎Step :- 2 }}$

$\tt \:  ⟼ \: \lim_{x\to 0}\dfrac{ {(1 + x)}^{5} - 1}{ {(1 + x)}^{3} - 1 }$

$\tt \:  put \: 1 + x = y \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \tt \: as \: x⟼0 \: then \: y⟼1$

$\tt \:  = \lim_{y\to 1}\dfrac{ {y}^{5} - 1 }{ {y}^{3} - 1 }$

☆ On dividing the numerator and denominator by y - 1, we get

$\tt \:  = \lim_{y\to 1}\dfrac{\dfrac{ {y}^{5} - 1}{y - 1} }{\dfrac{ {y}^{3} - 1}{y - 1} }$

$\tt \:  = \dfrac{5 \times {(1)}^{5 - 1} }{3 \times {(1)}^{2 - 1} }$

$\tt \:  = \dfrac{5}{3}$

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