Question

If $\displaystyle \lim_{x \to \infty} x^n \log_e x = 0$, then $\log_e 12 =$

• A. Negative
• B. Positive
• C. Zero
• D. any value between $-1$ and $1$

Hint:

A positive logarithm times a decaying polynomial can still lead to a vanishing limit, indicating the log must be comparatively small.

Answer 1

The Correct Option is A

Given $\displaystyle \lim_{x \to \infty} x^n \log_e x = 0$ implies that $n<0$, i.e., $x^n$ is a decaying function
Hence $\log_e x$ must be positive (since $x \to \infty$), but $x^n \log_e x \to 0$ only if $n<0$
Thus $\log_e 12>0$, but as $n<0$, the product $x^n \log_e x$ becomes a very small positive number
Hence, $\log_e 12$ must be negative to keep consistency with the limit expression going to 0