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Official PYQ · 2015Set 3 · Q9Calculus: Limits, Continuity, DifferentiabilityMCQ1 mark

GATE CS 2015 Set 3 Q9 — Calculus: Limits, Continuity, Differentiability

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The value of lim⁡x→∞(1+x2)e−x\displaystyle \lim_{x \rightarrow \infty} (1+x^2)^{e^{-x}} is

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Answer: C

To find the limit:

L=lim⁡x→∞(1+x2)e−xL = \lim_{x \rightarrow \infty} (1+x^2)^{e^{-x}}

This is an indeterminate form of the type ∞0\infty^0. Let us take the natural logarithm of both sides:

ln⁡L=lim⁡x→∞ln⁡((1+x2)e−x)\ln L = \lim_{x \rightarrow \infty} \ln\left((1+x^2)^{e^{-x}}\right) ln⁡L=lim⁡x→∞e−xln⁡(1+x2)=lim⁡x→∞ln⁡(1+x2)ex\ln L = \lim_{x \rightarrow \infty} e^{-x} \ln(1+x^2) = \lim_{x \rightarrow \infty} \frac{\ln(1+x^2)}{e^x}

As x→∞x \rightarrow \infty, this expression takes the indeterminate form ∞∞\frac{\infty}{\infty}. Applying L'Hôpital's Rule by differentiating the numerator and the denominator with respect to xx:

ln⁡L=lim⁡x→∞ddx[ln⁡(1+x2)]ddx[ex]\ln L = \lim_{x \rightarrow \infty} \frac{\frac{d}{dx}[\ln(1+x^2)]}{\frac{d}{dx}[e^x]} ln⁡L=lim⁡x→∞2x1+x2ex=lim⁡x→∞2xex(1+x2)\ln L = \lim_{x \rightarrow \infty} \frac{\frac{2x}{1+x^2}}{e^x} = \lim_{x \rightarrow \infty} \frac{2x}{e^x(1+x^2)}

Since the denominator grows exponentially (exe^x) and quadratically (1+x21+x^2), while the numerator grows only linearly (2x2x), the denominator dominates as x→∞x \rightarrow \infty:

ln⁡L=0\ln L = 0

Taking the exponential on both sides to solve for LL:

L=e0=1L = e^0 = 1

Therefore, the value of the limit is 1.

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