Exercise
Evaluate the following limit:
\lim _ { x \rightarrow 0} \frac{(e^x-1)(e^{2x}-1)}{x^2}
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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