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Calculating Limit of Function – A function with e in the power of a function to zero – Exercise 6319


Evaluate the following limit:

\lim _ { x \rightarrow 0} {(e^x+3x)}^{\frac{1}{x}}

Final Answer

\lim _ { x \rightarrow 0} {(e^x+3x)}^{\frac{1}{x}}=e^4


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}}=


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

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