Question

Calcule o seguinte limite sem utilizar L'Hospital:

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Answer 1

Explicação passo-a-passo:

$\displaystyle\lim_{\sf{x\to 0}}\left[\sf{\dfrac{e^x-1-x}{x^2}}\right]$

Note pela série de Maclaurin :

$\sf{e^x~=~}\displaystyle\sum\limits^{\infty}_{\sf{n~=~0}}\sf{\dfrac{x^n}{n!} }$

$\iff\sf{ e^x~=~1+x+\dfrac{x^2}{2}+\dfrac{x^3}{6}+\dfrac{x^4}{24}+\cdots }$

Substituindo no limite podemos ter :

$\iff \displaystyle\lim_{\sf{x \to 0}}\sf{\dfrac{\left(\cancel{\red{1}}\cancel{\pink{+x}}+\frac{x^2}{2}+\dfrac{x^3}{6}+\frac{x^4}{24}+\cdots\right)\cancel{\red{-1}}\cancel{\pink{-x}}}{x^2} }$

Ficamos assim com:

$\iff \displaystyle\lim_{\sf{x \to 0}}\sf{\dfrac{ \frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\cdots }{x^2}}$

Vamos agora evidenciar x² no númerador:

$\iff \displaystyle\lim_{\sf{x \to 0}}\sf{\dfrac{\cancel{ x^2}\left(\frac{1}{2}+\frac{x}{6}+\frac{x^2}{24}+\cdots\right)}{\cancel{x^2}}}$

$\iff \displaystyle\lim_{\sf{x \to 0}}\sf{\left[ \dfrac{1}{2}+\dfrac{x}{6}+\dfrac{x^2}{24}+\cdots \right]}$

$\iff \sf{~=~\dfrac{1}{2}+\dfrac{0}{6}+\dfrac{0}{24}+\cdots~=~\dfrac{1}{2} }$

$\iff \boxed{ \green{ \displaystyle\lim_{\sf{x \to 0}}\sf{\left[\dfrac{e^x-1-x}{x^2}\right]~=~\dfrac{1}{2}} } }$

This answer was elaborad by:

Murrima, Joaquim Marcelo

UEM(MOÇAMBIQUE)-DMI