Calculating Limit of Function – A quotient of exponential and polynomial functions to zero – Exercise 6303

Exercise

Evaluate the following limit:

$\lim _ { x \rightarrow 0} \frac{(e^x-1)(e^{2x}-1)}{x^2}$

Final Answer

Show final answer

Solution

First, we try to plug in $x = 0$ and get

$\frac{(e^0-1)(e^{2\cdot 0}-1)}{0^2}=\frac{0}{0}$

Note: The zero in the denominator is not absolute zero, but a number that is tending to zero.

We got the phrase $\frac{"0"}{"0"}$ (=tending to zero divides tending to zero). This is an indeterminate form, therefore we have to get out of this situation.

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

In order to use Lopital Rule, we open the brackets and get

$=\lim _ { x \rightarrow 0} \frac{e^{3x}-e^x-e^{2x}+1}{x^2}=$

Now we use Lopital Rule – we derive the numerator and denominator separately and we get

$=\lim _ { x \rightarrow 0} \frac{3e^{3x}-e^x-2e^{2x}}{2x}=$

We plug in zero again and get

$= \frac{3e^{3\cdot 0}-e^0-2e^{2\cdot 0}}{2\cdot 0}=$

$= \frac{0}{0}$

We got again $\frac{0}{0}$

So we we will use Lopital Rule again – we derive the numerator and denominator separately and we get

$=\lim _ { x \rightarrow 0} \frac{9e^{3x}-e^x-4e^{2x}}{2}=$

We plug in zero again and this time we get

$=\frac{9e^{3\cdot 0}-e^0-4e^{2\cdot 0}}{2}=$

$=\frac{9-1-4}{2}=$

$=\frac{4}{2}=$

$=2$

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