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  • Confused! kindly explain, - Limit , continuity and differentiability - JEE Main-10

\lim_{x\rightarrow 0}\frac{e^{x}-x-1}{x^{2}}  equals 

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best_answer

As we have learned

L - Hospital Rule -

In \:the\:form\:of\:\:\;\frac{0}{0}\:\:and\:\:\frac{\infty }{\infty }\:\:\:we\:differentiate\:\:\frac{N^{r}}{D^{r}}\:\:separately.


\Rightarrow \lim_{x\rightarrow a}\:\:\frac{f(x)}{g(x)}=\lim_{x\rightarrow a}\:\:\frac{f'(x)}{g'(x)}

- wherein

\lim_{x\rightarrow a}\:\:\frac{\frac{d}{dx}\:f(x)}{\frac{d}{dx}\:g(x)}


Where \:\:f(x)\:\:and\:\:g(x)=0

 

 The limit is of form 0/0 under x\rightarrow 0, so 

\lim_{x\rightarrow 0}\frac{e^{x}-x-1}{x^{2}}=\lim_{x\rightarrow 0}\frac{e^{x}-1}{2x}=1/2\lim_{x\rightarrow 0}\frac{e^{x}-1}{x}

=1/2*1=1/2

 

 

 

 


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