If the function \( f(x)=\begin{cases}\displaystyle\frac{e^{2\log_e x^2}-1}{x-1}, & x\ne 1\\a, & x=1\end{cases} \)
is continuous at \( x=1 \), then the value of \( a \) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 4
Since \( n\log_e m=\log_e m^n \) and \( e^{\log_e m}=m \),
\( \displaystyle e^{2\log_e x^2}=e^{\log_e x^4}=x^4 \)
\( \displaystyle \frac{x^4-1}{x-1}=\frac{(x-1)(x+1)(x^2+1)}{x-1} \)
\( \displaystyle =(x+1)(x^2+1),\quad x\ne 1 \)
For continuity at x = 1, a must equal the limit:
\( \displaystyle a=\lim_{x\to 1}(x+1)(x^2+1) \)
\( \displaystyle =2\times 2=4 \)
Shortcut: \( \lim_{x\to 1}\frac{x^n-1}{x-1}=n \), so with n = 4 the limit is 4 at once.
Why the other options are incorrect
Option (1) 1 — Incorrect. The limit is 4, not 1.
Option (2) 2 — Incorrect. 2 would come from \( e^{2\log_e x^2} \) read as \( x^2 \); the exponent 2 gives \( x^4 \).
Option (4) \( \log_e 2 \) — Incorrect. No logarithm remains after \( e^{\log_e x^4}=x^4 \); the limit is 4.