Question

find : Limit (1+X)^5 -1/3x+5x^2
x_0
with proper explanation​

Answer 1

To solve :-

$\rm\lim \limits_{x \to 0} \bigg\{\dfrac{(1+x)^5-1}{5x^2+3x} \bigg\}$

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Solution :-

Expanding (1+x)² using binomial theorem,

[see the expansion at : https://brainly.in/question/43788382 ] .

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${ \implies\rm\lim \limits_{x \to0} \bigg\{\dfrac{\big(x^5+5x^4+10x^3+10x^2+5x+1\big)-1}{5x^2+3x} \bigg\}}$

${\implies \rm\lim \limits_{x \to0} \bigg\{\dfrac{x^5+5x^4+10x^3+10x^2+5x+1-1}{x\big(5x+3\big)} \bigg\}}$

${\implies \rm\lim \limits_{x \to0} \bigg\{\dfrac{x^5+5x^4+10x^3+10x^2+5x}{x\big(5x+3\big)} \bigg\}}$

${ \implies \rm\lim \limits_{x \to0} \bigg\{\dfrac{x \big(x^4+5x^3+10x^2+10x+5 \big)}{x\big(5x+3\big)} \bigg\}}$

${ \implies\rm\lim \limits_{x \to0} \bigg\{\dfrac{ \not x \big(x^4+5x^3+10x^2+10x+5 \big)}{ \not x\big(5x+3\big)} \bigg\}}$

${\implies \rm\lim \limits_{x \to0} \bigg\{\dfrac{ x^4+5x^3+10x^2+10x+5 }{ 5x+3} \bigg\}}$

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Now directly substitute x = 0

${ \rm\implies\bigg\{\dfrac{ 0^4+5(0)^3+10(0)^2+10(0)+5 }{ 5(0)+3} \bigg\}}$

${\rm\implies\bigg\{\dfrac{ 5 }{ 3} \bigg\}}$

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Hence,

$\rm\lim \limits_{x \to 0} \bigg\{\dfrac{(1+x)^5-1}{5x^2+3x} \bigg\} =\bigg\{\dfrac{ 5 }{ 3} \bigg\}$