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Official PYQ · 2015Set 1 · Q4Calculus: Limits, Continuity, DifferentiabilityMCQ1 mark

GATE CS 2015 Set 1 Q4 — Calculus: Limits, Continuity, Differentiability

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lim⁡x→∞x1x\displaystyle \lim_{x\rightarrow \infty } x^{ \tfrac{1}{x}} is

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Answer: C

To find the limit L=lim⁡x→∞x1xL = \lim_{x\rightarrow \infty } x^{\frac{1}{x}}, we can use logarithms since it is of the indeterminate form ∞0\infty^0.

Let y=x1xy = x^{\frac{1}{x}}. Taking the natural logarithm on both sides:

ln⁡(y)=ln⁡(x1x)=ln⁡(x)x\ln(y) = \ln\left(x^{\frac{1}{x}}\right) = \frac{\ln(x)}{x}

Now, take the limit as x→∞x \to \infty:

lim⁡x→∞ln⁡(y)=lim⁡x→∞ln⁡(x)x\lim_{x\to\infty} \ln(y) = \lim_{x\to\infty} \frac{\ln(x)}{x}

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

lim⁡x→∞ln⁡(x)x=lim⁡x→∞ddx(ln⁡(x))ddx(x)=lim⁡x→∞1/x1=0\lim_{x\to\infty} \frac{\ln(x)}{x} = \lim_{x\to\infty} \frac{\frac{d}{dx}(\ln(x))}{\frac{d}{dx}(x)} = \lim_{x\to\infty} \frac{1/x}{1} = 0

Since lim⁡x→∞ln⁡(y)=0\lim_{x\to\infty} \ln(y) = 0, we have:

lim⁡x→∞y=e0=1\lim_{x\to\infty} y = e^0 = 1

Thus, lim⁡x→∞x1x=1\lim_{x\rightarrow \infty } x^{\frac{1}{x}} = 1.

Therefore, the correct option is C.

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