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  • I have a doubt, kindly clarify. - Limit , continuity and differentiability - JEE Main-15

\lim_{x\rightarrow 0}\frac{x^{2}}{e^{\left [ x \right ]}} equals

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Answers (1)

best_answer

As we have learned

The SANDWICH THEOREM -

If \:\:f(x)\leq g(x)\leq h(x)\:\:for\:every\:(x)\:in\:the\:deleted\:neighbourhood\:of\:(a).


and\:\:\:\lim_{x\rightarrow a}\:\:f(x)=\lim_{x\rightarrow a}\:\:h(x)=l


Then\;\:\lim_{x\rightarrow a}\:g(x)\:\:is\:also\:equal\:to\:l

- wherein

 

Where\:\:\:x\epsilon \:(a-\delta ,\:a+\delta )\:\:and\:\:\delta\: \:is\:very\:small.

 

 e^{x-1}< e^{[x]} \Rightarrow \frac{x^{2}}{e^{x}-1}> \frac{x^{2}}{e^{[x]}}> \frac{x^{2}}{e^{x}}

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

and  \lim_{x\rightarrow \infty } \frac{x^{2}}{e^{x}} =\lim_{x\rightarrow \infty }\frac{2x}{e^{x}}=\lim_{x\rightarrow \infty } \frac{2}{e^{x}}=0

(Both limits calculated using L-HOSPITAL )

 

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

\therefore \lim_{x\rightarrow \infty } \frac{x^{2}}{e^{[x]}} =0  (using SANDWICH THEOREM)

 

 

 


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Posted by

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