# Calculating Limit of Function – ln function multiplies by ln function to infinity – Exercise 6587

### Exercise

Evaluate the following limit:

$$\lim _ { x \rightarrow \infty} \ln (1+2^x)\ln (1+\frac{3}{x})$$

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

Solution

First, we try to plug in $$x = \infty$$ and get

$$\ln (1+2^\infty)\ln (1+\frac{3}{\infty})=\infty\cdot 0$$

We got the phrase $$0\cdot\infty$$ (=tends to zero multiplies by infinity). This is an indeterminate form, therefore we have to get out of this situation.

Using Logarithm Rules we get

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

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

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

Note: One can insert the boundary because ln function is a continuous function.

We will use the known limit also called Euler’s Limit. At the base, we have an expression of the form:

$$1+\frac{3}{x}$$

And the following holds

$$\lim _ { x \rightarrow \infty}\frac{3}{x}= 0$$

As required.

Therefore, we will multiply the power by the inverted expression and get

$$=\ln \lim _ { x \rightarrow \infty} {(1+\frac{3}{x})}^{\frac{x}{3}\cdot\frac{3}{x}\cdot\ln (1+2^x)}=$$

Note: When adding an expression by multiplication, its inverted expression must also be multiplied so that the original expression does not change.

Now, according to Euler’s Limit, we get

$$\lim _ { x \rightarrow \infty} {(1+\frac{3}{x})}^{\frac{x}{3}}=e$$

All we have left to do is to calculate the limit on the expression that remains in the power:

$$\lim _ { x \rightarrow \infty}\frac{3}{x}\cdot\ln (1+2^x)=$$

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

We plug in infinity and get

$$=\frac{3\ln (1+2^\infty)}{\infty}=\frac{\infty}{\infty}$$

We got the phrase $$\frac{\infty}{\infty}$$. This is also an indeterminate form, therefore we use Lopital Rule – we derive the numerator and denominator separately and we will get

$$=3\lim _ { x \rightarrow \infty}\frac{\frac{1}{1+2^x}\cdot 2^x\cdot\ln 2}{1}=$$

$$=3\ln 2\lim _ { x \rightarrow \infty}\frac{2^x}{1+2^x}=$$

Plugging in infinity gives the phrase $$\frac{\infty}{\infty}$$ again. This time we will divide both numerator and denominator by $$2^x$$ and get

$$=3\ln 2\lim _ { x \rightarrow \infty}\frac{1}{\frac{1}{2^x}+1}=$$

We plug in infinity again and this time we get

$$=3\ln 2\cdot\frac{1}{\frac{1}{2^\infty}+1}=$$

$$=3\ln 2\cdot\frac{1}{\frac{1}{\infty}+1}=$$

$$=3\ln 2\cdot\frac{1}{0+1}=$$

$$=3\ln 2$$

Therefore, in total we get

$$=\ln \lim _ { x \rightarrow \infty} {(1+\frac{3}{x})}^{\frac{x}{3}\cdot\frac{3}{x}\cdot\ln (1+2^x)}=$$

$$=\ln e^{3\ln 2}=$$

$$=3\ln 2$$

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