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  • Give answer! - Limit , continuity and differentiability - JEE Main-6

\lim_{x\rightarrow \infty }\frac{[x]}{e^{x}} equals

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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.

 

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

\lim_{x\rightarrow \infty }\frac{x-1}{e^{x}} = \lim_{x\rightarrow \infty } 1/e^{x}=0 ( using L hospital rule)

\lim_{x\rightarrow \infty } \frac{x}{e^{x}}= \lim_{x\rightarrow \infty } 1/e^{x}=0  (again using L-hospital rule as \infty /\infty  form )

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

\therefore using sandwich theorem  

\lim_{x\rightarrow \infty } \frac{[x]}{e^{x}}=0

 

 

 

 

 

 

 

 


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