1. ## calculus - limits

evaluate:

lim x approach infinity [sqrt(e^2x-e^x)-e^x]

tried alotta things, didn work, help pls!

2. Originally Posted by ssdimensionss
evaluate:

lim x approach infinity [sqrt(e^2x-e^x)-e^x]

tried alotta things, didn work, help pls!
$= \lim_{x \rightarrow + \infty}\frac{\sqrt{(e^{2x} - e^x} - e^x)(\sqrt{e^{2x} - e^x} + e^x)}{\sqrt{e^{2x} - e^x} + e^x}$

$= \lim_{x \rightarrow + \infty}\frac{e^{2x} - e^x - e^{2x}}{\sqrt{e^{2x} - e^x} + e^x}$

$= \lim_{x \rightarrow + \infty} - \frac{e^x}{\sqrt{e^{2x} - e^x} + e^x}$

$= \lim_{x \rightarrow + \infty} - \frac{1}{\sqrt{1 - e^{-x}} + 1}$

$= - \frac{1}{2}$.

3. ## thanks

hey thx for the help! i did it all up to the second to last step, the bit i dont get is why does e^x limit for x apporach infinty = 1?? please clarify, thx again!

4. Originally Posted by ssdimensionss
hey thx for the help! i did it all up to the second to last step, the bit i dont get is why does e^x limit for x apporach infinty = 1?? please clarify, thx again!
$e^{-x} \rightarrow 0$ as $x \rightarrow +\infty$. So you could insert the following line between the second last and last:

$= - \frac{1}{\sqrt{1 - 0} + 1}$

Capisce?

5. ## ^

yes but that still does not explain why the numerator which was lim x approach infinity e^x is = 1. sorry for bothering again!

6. Hello,

He simplified the numerator and the denominator with e^x, so e^x becomes 1.

7. Originally Posted by ssdimensionss
yes but that still does not explain why the numerator which was lim x approach infinity e^x is = 1. sorry for bothering again!
I divided the numerator and denominator by $e^x$. When you divide the numerator by $e^x$ you get 1.

Note on denominator: When dividing outside the square root by $e^x$, you divide inside the square root by $e^{2x}$:

$\frac{\sqrt{e^{2x} - e^x}}{e^x} = \sqrt{\frac{e^{2x} - e^x}{e^{2x}}} = \sqrt{1 - e^{-x}}$.

8. i understand now thx so much!

9. i understand now thx so much!