Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Continuity and Differentiability

Question:

Evaluate: $\underset{x→e}{\lim}\frac{\log x-1}{x-e}$.

Options:

e

-e

$\frac{1}{e}$

$-\frac{1}{e}$

Correct Answer:

$\frac{1}{e}$

Explanation:

$\underset{x→e}{\lim}\frac{ln\, x-1}{x-e}=\underset{x→e}{\lim}\frac{ln\,x-ln\,e}{x-e}=\underset{x→e}{\lim}\frac{ln(x/e)}{x-e}=\underset{x→e}{\lim}\frac{ln(1+\frac{x-e}{e})}{\frac{x-e}{e}.e}$

$=\underset{x→e}{\lim}\frac{ln(1+\frac{x-e}{e})^{e/x-e}}{e}=\frac{1}{e}$