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

### Exercise

Evaluate the following limit:

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

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

### 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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