Exercise
Evaluate the following limit:
\lim _ { x \rightarrow 0} {(e^x+3x)}^{\frac{1}{x}}
Final Answer
Solution
First, we try to plug in x = 0 and get
{(e^0+3\cdot 0)}^{\frac{1}{0}}
We got the phrase 1^0 (=tending to 1 in the power of tending to 0). This is an indeterminate form, therefore we have to get out of this situation.
\lim _ { x \rightarrow 0} {(e^x+3x)}^{\frac{1}{x}}=
Using Logarithm Rules we get
=\lim _ { x \rightarrow 0} e^{\ln {(e^x+3x)}^{\frac{1}{x}}}=
=\lim _ { x \rightarrow 0} e^{\frac{1}{x}\ln (e^x+3x)}=
We simplify the phrase and get
=\lim _ { x \rightarrow 0} e^{\frac{\ln (e^x+3x)}{x}}=
We enter the limit inside
= e^{\lim _ { x \rightarrow 0} \frac{\ln (e^x+3x)}{x}}=
Note: This can be done because an exponential function is a continuous function.
We plug in 0 again and get
=e^{\frac{\ln (e^0+3\cdot 0)}{0}}=
=e^{\frac{\ln 1}{0}}=
=e^{\frac{0}{0}}=
We got the phrase \frac{"0"}{"0"}(=tending to zero divides tending to zero). This is also an indeterminate form, in such cases we use Lopital Rule – we derive the numerator and denominator separately and we will get
= e^{\lim _ { x \rightarrow 0} \frac{\frac{1}{e^x+3x}\cdot (e^x+3)}{1}}=
We simplify the phrase and get
= e^{\lim _ { x \rightarrow 0} \frac{e^x+3}{e^x+3x}}=
We plug in zero again and get
= e^{ \frac{e^0+3}{e^0+3\cdot 0}}=
= e^{ \frac{1+3}{1+0}}=
= e^4
Have a question? Found a mistake? – Write a comment below!
Was it helpful? You can buy me a cup of coffee here, which will make me very happy and will help me upload more solutions!