Question

please answer this question with detailed solution.

A)0
B)1
C)-1
D)2

Answer 1

$\Large{\underline{\underline{\mathfrak{\bf{Solution}}}}}$

$\Large{\underline{\mathfrak{\bf{Find}}}}$

• $\sf{\blue{\displaystyle \lim_{x \to 0 } \dfrac{e^x+e^{-x}-2}{x^2}}}$.

$\Large{\underline{\underline{\mathfrak{\bf{Explanation}}}}}$

$:\mapsto\sf{\red{\:\displaystyle\lim_{x \to 0 } \dfrac{e^x+e^{-x}-2}{x^2}}} \\ \\ \\ :\mapsto\sf{\:\displaystyle \lim_{ x \to 0} \dfrac{e^x+\dfrac{1}{e^x}-2}{x^2}} \\ \\ \\ :\mapsto\sf{\:\displaystyle \lim_{x \to 0} \dfrac{e^x.e^{x}+1-2e^{-x}}{x^2}} \\ \\ \\ :\mapsto\sf{\:\displaystyle \lim_{x \to 0} \dfrac{e^{2x}+1-2e^x}{x^2}} \\ \\ \\ :\mapsto\sf{\:\displaystyle \lim_{x \to 0} \dfrac{(e^x-1)^2}{x^2}} \\ \\ \\ :\mapsto\sf{\:\displaystyle \lim_{x \to 0} \left(\dfrac{e^x-1}{x^2}\right)^2}$

But, we know,

$\bigstar\red{\:\displaystyle \lim_{x \to 0} \dfrac{e^x-1}{x}\:=\:1}$

So,

$:\mapsto\sf{\:(1)^2} \\ \\ \\ :\mapsto\sf{\pink{\:1\:\:\:\:\:Ans.}}$

$\Large{\underline{\mathfrak{\bf{Thus}}}}$

• Your answer will be options number (B).

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