calculus - limits

**ssdimensionss** · March 23rd 2008, 03:01 AM

evaluate:

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

tried alotta things, didn work, help pls!

**mr fantastic** · March 23rd 2008, 03:15 AM

Originally Posted by ssdimensionss

evaluate:

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

tried alotta things, didn work, help pls!

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

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

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

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

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

**ssdimensionss** · March 23rd 2008, 04:09 AM

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!

**mr fantastic** · March 23rd 2008, 04:43 AM

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?

**ssdimensionss** · March 23rd 2008, 05:25 AM

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

**Moo** · March 23rd 2008, 05:27 AM

Hello,

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

**mr fantastic** · March 23rd 2008, 02:30 PM

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 $e^{2x}$ :

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

**ssdimensionss** · March 27th 2008, 02:34 AM

i understand now

thx so much!

**ssdimensionss** · March 27th 2008, 03:29 AM

i understand now

thx so much!

Thread: calculus - limits

LinkBack

Thread Tools

Display

calculus - limits

thanks

^

Similar Math Help Forum Discussions

Calculus 2 limits as x-->0+

Calculus, limits

Calculus, limits

Calculus Limits Help

calculus limits

Search Tags