# Calculating Limit of Function – A quotient of exponential functions to infinity – Exercise 6556

### Exercise

Evaluate the following limit:

$$\lim _ { x \rightarrow \infty} \frac{4^{-x}-5^{-x}}{2^{-x}-3^{-x}}$$

$$\lim _ { x \rightarrow \infty} \frac{4^{-x}-5^{-x}}{2^{-x}-3^{-x}}=0$$

### Solution

First, we try to plug in

$$x =\infty$$

and get

$$\frac{4^{-\infty}-5^{-\infty}}{2^{-\infty}-3^{-\infty}}=\frac{0-0}{0-0}$$

We got the phrase 0/0 (=tends to zero divides by tends to zero). This is an indeterminate form, therefore we have to get out of this situation.

Since we have a quotient of exponential functions going to infinity, we divide the numerator and denominator by the expression that aspires most rapidly to infinity, without its coefficient. This division gives us the following:

$$\lim _ { x \rightarrow \infty} \frac{2^{-x}(2^{-x}-{(\frac{5}{2})}^{-x})}{2^{-x}(1-{(\frac{3}{2})}^{-x})}=$$

$$=\lim _ { x \rightarrow \infty} \frac{2^{-x}-{(\frac{5}{2})}^{-x}}{1-{(\frac{3}{2})}^{-x}}=$$

$$=\lim _ { x \rightarrow \infty} \frac{{(\frac{1}{2})}^{x}-{(\frac{2}{5})}^{x}}{1-{(\frac{2}{3})}^{x}}=$$

Since the base is less than 1, the following holds:

$$=\lim _ { x \rightarrow \infty} {(\frac{2}{3})}^x=0$$

$$=\lim _ { x \rightarrow \infty} {(\frac{2}{5})}^x=0$$

$$=\lim _ { x \rightarrow \infty} {(\frac{1}{2})}^x=0$$

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We plug in infinity again, and this time we get

$$=\frac{0-0}{1-0}=$$

$$=\frac{0}{1}=$$

$$=0$$

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