Question

If
\[ \lim_{x\to\infty}x^{\log_e x}=0, \] then \(\log_x12=\)

• A. Negative
• B. Positive
• C. Zero
• D. Any value between \(-1\) and \(1\)

Hint:

For logarithms, \(\log_a b\) is negative whenever \(0\lt a\lt 1\) and \(b\gt 1\).

Answer 1

The Correct Option is A

Step 1: Interpret the given limit.
Given,
\[ \lim_{x\to\infty}x^{\log_e x}=0 \]
Using the identity
\[ a^{\log b}=e^{(\log a)(\log b)}, \] we write
\[ x^{\log_e x}=e^{(\log_e x)^2} \]
Thus,
\[ \lim_{x\to\infty}e^{(\log_e x)^2}=0 \]
For an exponential expression \(e^t\) to approach \(0\), the exponent must tend to \(-\infty\).
Hence,
\[ (\log_e x)^2\to -\infty \] which is possible only when
\[ \log_e x\lt 0 \]
Therefore,
\[ 0\lt x\lt 1 \]

Step 2: Determine the sign of \(\log_x12\).
We know that
\[ \log_x12=\frac{\log_e12}{\log_e x} \]
Since \(12\gt 1\),
\[ \log_e12\gt 0 \]
Also, from Step 1,
\[ \log_e x\lt 0 \]
Hence,
\[ \frac{\text{positive}}{\text{negative}}\lt 0 \]
Therefore,
\[ \log_x12\lt 0 \]

Step 3: Final conclusion.
Hence,
\[ \boxed{\text{Negative}} \]