Troubleshooting Indeterminate Limits in Calculus

[phrox]

http://i1301.photobucket.com/albums/ag115/phrox1/additional_zps10bf2373.png

That's the question ^^,

So I came up with H = sqrt(D^2 + 1)

Then I got

H - D = (sqrt(1/x^2)+1) - D

Annndd then:

H - 1/x = (sqrt(1/x^2)+1) - sqrt(1/x^2)

I have no clue where to go from here, any help is appreciated!

 

[phrox]

So did is my equation for a) H - 1/x = (sqrt(1/x^2)+1) - sqrt(1/x^2)

and for b) it says instead, compute the limit as x-> 0, which would make it not work because all the x's are under ones, making them undefined.

 

[topsquark]

So did is my equation for a) H - 1/x = (sqrt(1/x^2)+1) - sqrt(1/x^2)

and for b) it says instead, compute the limit as x-> 0, which would make it not work because all the x's are under ones, making them undefined.

So far so good. Have you covered how to use L'Hopital's rule for the indeterminant form \infty - \infty?

-Dan

 

[phrox]

no I haven't covered L'Hospital's rule, also I can't directly sub x->0 into this because it just won't work, undefined.

 

[topsquark]

Well, okay. I'll leave the final word to someone who knows limits better than me, but here's a "hand waving" argument.

For small x, 1/x^2 gets very large. So large that we can ignore the 1 in the first expression:
[math]\frac{1}{x^2} + 1 \to \frac{1}{x^2}[/math]

so
[math]\lim_{x \to 0} \left ( \sqrt{\frac{1}{x^2} + 1} - \sqrt{\frac{1}{x^2}} \right ) \to \sqrt{\frac{1}{x^2}} - \sqrt{\frac{1}{x^2}} = 0[/math]

-Dan

 

[chisigma]

So did is my equation for a) H - 1/x = (sqrt(1/x^2)+1) - sqrt(1/x^2)

and for b) it says instead, compute the limit as x-> 0, which would make it not work because all the x's are under ones, making them undefined.

Very well!... what You have to do is to evaluate...

$\displaystyle l= \lim_{x \rightarrow 0} \sqrt{1 + \frac{1}{x^{2}}} - \sqrt{\frac{1}{x^{2}}}\ (1)$

Now You can consider that...

$\displaystyle \sqrt{1 + \frac{1}{x^{2}}} - \sqrt{\frac{1}{x^{2}}} = (\sqrt{1 + \frac{1}{x^{2}}} - \sqrt{\frac{1}{x^{2}}})\ \frac{\sqrt{1 + \frac{1}{x^{2}}} + \sqrt{\frac{1}{x^{2}}}}{\sqrt{1 + \frac{1}{x^{2}}} + \sqrt{\frac{1}{x^{2}}}}= \frac{1}{\sqrt{1 + \frac{1}{x^{2}}} + \sqrt{\frac{1}{x^{2}}}}\ (2)$

... and the 'indeterminate form' disappared...

Kind regards

$\chi$ $\sigma$

 

[alyafey22]

[math]\sqrt{\frac{1}{x^2}} - \sqrt{\frac{1}{x^2}} = 0[/math]

-Dan

You can not remove the limit because it is still indeterminate form . You can not simply cancel .

 

[phrox]

I'm guessing you can't do it simply this way? http://s1301.photobucket.com/user/phrox1/media/image001_zps75e88aa9.jpg.html?sort=3&o=0

 

[Opalg]

I'm guessing you can't do it simply this way? http://s1301.photobucket.com/user/phrox1/media/image001_zps75e88aa9.jpg.html?sort=3&o=0

No you can't, because the expression $\infty - \infty$ is not defined. What you can do is to use the trick in chisigma's comment above, to write the limit as $$\lim_{x\to0}\left( \frac{1}{\sqrt{1 + \frac{1}{x^{2}}} + \sqrt{\frac{1}{x^{2}}}}\right).$$ Multiply top and bottom of that fraction by $x$, so that it becomes $$\lim_{x\to0}\left( \frac{x}{\sqrt{x^2 + 1} + 1}\right).$$ In that fraction, the numerator goes to $0$ as $x\to0$, but the denominator goes to $2$. So the fraction has limit $\dfrac02 = 0.$

 

[phrox]

Thanks a ton, I just find it so weird thinking of x as in sqrt(x^2) and etc. Wish I had time to practice limits instead of getting an assignment as soon as I'm finished the last lol.

 

[topsquark]

You can not remove the limit because it is still indeterminate form . You can not simply cancel .

I really should have written it as
[math]\lim_{x \to 0} \left ( \sqrt{ \frac{1}{x^2} } - \sqrt{ \frac{1}{x^2} } \right )[/math]

but I did say it was "hand waving"...

-Dan