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

e−∞ is equal to zero. can we directly substitute infinity on x it is giving zero any thing power zero is 1 am i right ?

 

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C is Correct

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Ans: C. 

Answer 1

Apply an exponential of a logarithm to the expression.

$\quad\lim\limits_{x\to \infty}(x^{2}+1)^{e^{-x}}=\lim\limits_{x\to \infty} \exp \left(\log ((x^{2}+1)^{e^{-x}})\right)$

$=\lim\limits_{x\to \infty} \exp \left(\dfrac{\log(x^{2}+1)}{e^{x}}\right)$

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Since the exponential function is continuous, we may factor it out of the limit.

$\quad\lim\limits_{x\to \infty} \exp \left(\dfrac{\log(x^{2}+1)}{e^{x}}\right)$

$=\exp \left(\lim\limits_{x\to \infty} \dfrac{\log(x^{2}+1)}{e^{x}}\right)$

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The numerator of $e^{-x} \log (x^{2}+1)$  grows asymptotically slower than its denominator as $x$ approaches $\infty$.

Since $\log (x^{2}+1)$ grows asymptotically slower than $e^{x}$ as $x$ approaches $\infty$, $\lim\limits_{x\to \infty}e^{-x} \log (x^{2}+1)=0 : e^{0}.$

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Evaluate $e^{0}$.

$e^{0}=1$:

Answer: $1.$

Correct Answer: $C$

Comments

its answer is wrong i think

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What is wrong with this answer?

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its 1/(1+x^2)^e^x. so simple subsitituion gives value 1/infinity=0;

Answer 2

$\lim_{x \to \infty}(1+x^2)^{e^-x}=y(let)$

Take log

$log(y)=e^{-x}.log(1+x^2)=lim_{x \to \infty} \frac{log(1+x^2)}{e^x}=\frac{\infty}{\infty}form$

Apply L Hospital rule

$lim_{x \to \infty} \frac{\frac{2x}{1+x^2}}{e^x}=\frac{2x}{e^x(1+x^2)}$

Differentiate one-more time

$\lim_{x \to \infty} \frac{2}{e^x(1+x^2)+2xe^x}=0$

$log(y)=0 \rightarrow y=1$

Answer : (1)

Tip: For power functions it always better to take log and solve.

Comments

I meant the limit value.
limx→∞ 2/ex(1+x2)+2xex=0 how is this zero

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denominator becomes infinity as e^infinity is infinity. 2/infinity is 0.

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Oh ya got it thanks...

Answer 3

Here the base of the log is assumed to be e(Natural log).

Hence Y=e⁰ = 1 is ans.(The last step)

Comments

Is this way correct?
$log(y) = e^{-x}[log(1)\times 2log(x)]$
But $log(1)=0$, therefore
$log(y)=0$
taking exponentiation
$y=e^0=1$
$\lim_{x\rightarrow\infty} 1 = 1$

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@habedo007 It is correct because we haven't used limits, which gives $0*\infty$ form, creating confusion.

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Those who didn't know about Indeterminate forms may refer to the link below, it's worth looking:

https://brilliant.org/wiki/indeterminate-forms/

Answer 4

I think this would help you to solve this type of question in much easier way 

Comments

How are you getting 1^infinity to apply this formula>?

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I don't think it is correct, because here e^-x is given which tends to '0' ,not infinity,but you assume it as infinity.

This formula is for 1^infinity.So, this is wrong.

Answer 5

  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

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 6

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