MCAPREP
Home > PYQ Blogs > CUET-MCA 2025 Paper > Question 60

Question 60

Mathematics Limit of Functions Easy

The value of \( lim_{x\rightarrow\infty}(1+\frac{2}{3x})^{x}\) is

(A) e
(B) \( e^{2}\)
(C) \(e^{\frac{2}{3}}\)
(D) \( e^{3}\)
View Dynamic Solution & Explanation ▼
Correct Solution: Option C

Step-by-step Solution:

Evaluate the limit

Problem:

\[ \lim_{x\to\infty}\left(1+\frac{2}{3x}\right)^{x} \]

Step 1 — Recognize the standard limit form.
We use the well-known limit \[ \lim_{t\to\infty}\left(1+\frac{a}{t}\right)^{t}=e^{a}, \] valid for any constant \(a\).
Step 2 — Match the expression to the standard form.
Rewrite the given expression as \[ \left(1+\frac{2}{3x}\right)^{x} =\left(1+\frac{a}{x}\right)^{x} \] with \(a=\dfrac{2}{3}\).
Step 3 — Apply the standard limit.
Therefore \[ \lim_{x\to\infty}\left(1+\frac{2}{3x}\right)^{x} = e^{\,\frac{2}{3}}. \]
Final Answer: \(\displaystyle e^{\frac{2}{3}}\) (Option C)

Remark: another tiny justification is to set \(t=x\) and directly apply the identity, or write \(\left(1+\dfrac{2}{3x}\right)^{x}=\Big[\left(1+\dfrac{2}{3x}\right)^{3x/2}\Big]^{2/3}\) and use \(\lim_{u\to\infty}\big(1+\tfrac{1}{u}\big)^{u}=e\).

← Previous Question Our Rankers Next Question →