Question

$\underset{x\to \infty }{Lt}x\left(a^{\frac{1}{x}}-b^{\frac{1}{x}}\right)=$

• A. 1
• B. ${\log }_{e}a/b$
• C. ${\log }_e\left(ab\right)$
• D. 0

Answer 1

The correct option is B ${\log }_{e}a/b$

Now, to evaluate,

$\underset{x\to \infty }{Lt}x\left(a^{\frac{1}{x}}-b^{\frac{1}{x}}\right)$......(1).

Let $y=\frac{1}{x}$.

Then as $x\to \infty$, $y\to 0$.

So the above expression (1) becomes,

$\underset{y\to 0}{Lt}\frac{\left(a^y-b^y\right)}{y}$

$=\underset{y\to 0}{Lt}\frac{\left(\right\{a^y-1\}-\{b^y-1\left\}\right)}{y}$

$=\underset{y\to 0}{lim}(\frac{a^y-1}{y}-\frac{b^y-1}{y})$

$={\log }_{e}a-{\log }_{e}b$

$={\log }_e\frac{a}{b}$.