GATE CSE 2015 Set 3 | Question: 9

Question

The value of $\displaystyle \lim_{x \rightarrow \infty} (1+x^2)^{e^{-x}}$ is

• A. $0$
• B. $\frac{1}{2}$
• C. $1$
• D. $\infty$

Comments

(1+x ^2) * e^-x = (1+x ^2) /e ^ x so its infinity/infinity form so on applyig L-hospitals rule ,

differnetiating Num & denominator seperately

we get                       =  2^x / e^x   diif again , = 2/e^x

on putting limits          = 2/ infinity = 0    so answer (A) = 0 is correct

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It is $\lim_{x \rightarrow \infty} (1+x^2)^{e^{-x}}$ not $\lim_{x \rightarrow \infty} (1+x^2)*{e^{-x}}$
Check http://www.gate.iisc.ernet.in/gate-ques-2015/CS_SET_3_GATE_2015.pdf Q14
and answer is C  http://www.gate.iisc.ernet.in/gate-answer-2015/CS_S07.pdf Q24

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ohh.. thaku for pointing it out

Answer 1

  lim->infinity(1+x²)/e^x when we put infinity in given expression then it will give infinity /infinity form and in infinity/infinity form we apply L hospital rule by  applying L hospital rule it two times given function becomes

2/e^x 

now we calculate limx->infinty (2/e^x)  = 0 

Comments

1 is correct answer !

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Why is option C wrong? I am not getting, both seem like right :( which method is right??

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For applying L'Hospitals's rule we need to have 0/0 form. But it is in infinity/infinity form. So we can't apply that rule. 1 is correct answer.

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we can use LH for 0/0.

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yes we can use it in only 0/0 form not in any other form.

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we can use LH on infi/infi too...

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Yes you are right
Ref : http://tutorial.math.lamar.edu/Classes/CalcI/LHospitalsRule.aspx

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@saket nandan plrase check @ਜਗਮੀਤ's top comment.

Answer 2

Let log y= e^(-x)* log(1+x^2)

               =e^(-x)*log 1*log x^2 =0 ; log 1 =0

log y =0

y=e^0=1