Question

Evaluate:
$\rm \displaystyle \lim_{x \to 0}\ \frac{9^{x}-5^{x}}{4^{x}-1}$

Answer 1

❤see in the given pic

Answer 2

Answer:

,$\rm \displaystyle \lim_{x \to 0}\ \frac{9^{x}-5^{x}}{4^{x}-1} = \frac{log\frac{9}{5} }{log4}$

Step-by-step explanation:

SInce we need to evaluate the $\rm \displaystyle \lim_{x \to 0}\ \frac{9^{x}-5^{x}}{4^{x}-1}$

Apply directly $\lim_{x\to 0}$ we get ,

$\frac{9^{0}-5^{0}}{4^{0}-1 } =\frac{1-1}{1-1}= \frac{0}{0}$

We know that $9^{0} = 1, 5^{0} =1$ and $4^{0} = 1$

So, apply L'Hospital rule (Differentiate the function with respect to its variable x ) we get,

We get $\frac{9^{x}log9-5^{x}log5 }{4^{x}log4}$

Apply limit we get

$\frac{log9-log5}{log4} = \frac{log\frac{9}{5} }{log4}$

So,$\rm \displaystyle \lim_{x \to 0}\ \frac{9^{x}-5^{x}}{4^{x}-1} = \frac{log\frac{9}{5} }{log4}$