LimitsmediumFree

Limits — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question

The value of

\displaystyle\lim_{x\to\infty}\dfrac{(2x^{n})^{1/e^{x}}-(3x^{n})^{1/e^{x}}}{x^{n}}

(where

n\in\mathbb{N}

) is

\ln\dfrac{2}{3}

0

correct

n\ln\dfrac{2}{3}

DNot defined

Solution

Step 1: Rewrite each term. As

x\to\infty

\dfrac{1}{e^{x}}\to 0

, so

(2x^{n})^{1/e^{x}}=e^{(\ln 2+n\ln x)/e^{x}}

. Step 2: For small exponent

u

e^{u}\sim 1+u

. So:

(2x^{n})^{1/e^{x}}\sim 1+\dfrac{\ln 2+n\ln x}{e^{x}}

(3x^{n})^{1/e^{x}}\sim 1+\dfrac{\ln 3+n\ln x}{e^{x}}

Step 3: Subtract:

(2x^{n})^{1/e^{x}}-(3x^{n})^{1/e^{x}}\sim\dfrac{\ln 2-\ln 3}{e^{x}}=\dfrac{\ln(2/3)}{e^{x}}

Step 4: Divide by

x^{n}

\lim_{x\to\infty}\dfrac{\ln(2/3)}{x^{n}\cdot e^{x}}=0

(since

x^{n}\cdot e^{x}\to\infty

). Correct answer: (2)

0

Still stuck on this question?Ask your doubt on WhatsApp

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.