Exercise
Evaluate the following limit:
\lim _ { x \rightarrow \infty} \ln (1+2^x)\ln (1+\frac{3}{x})
Final Answer
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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