- Thread starter
- #1

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!

- Thread starter phrox
- Start date

- Thread starter
- #1

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!

- Thread starter
- #2

- Aug 30, 2012

- 1,199

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

-Dan

- Thread starter
- #4

- Aug 30, 2012

- 1,199

For small x, 1/x^2 gets very large. So large that we can ignore the 1 in the first expression:

\(\displaystyle \frac{1}{x^2} + 1 \to \frac{1}{x^2}\)

so

\(\displaystyle \lim_{x \to 0} \left ( \sqrt{\frac{1}{x^2} + 1} - \sqrt{\frac{1}{x^2}} \right ) \to \sqrt{\frac{1}{x^2}} - \sqrt{\frac{1}{x^2}} = 0\)

-Dan

- Feb 13, 2012

- 1,704

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

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

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

$\chi$ $\sigma$

- Jan 17, 2013

- 1,667

You can not remove the limit because it is still indeterminate form . You can not simply cancel .\(\displaystyle \sqrt{\frac{1}{x^2}} - \sqrt{\frac{1}{x^2}} = 0\)

-Dan

- Thread starter
- #8

I'm guessing you can't do it simply this way? image001_zps75e88aa9.jpg Photo by phrox1 | Photobucket

- Moderator
- #9

- Feb 7, 2012

- 2,785

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 \(\displaystyle \lim_{x\to0}\left( \frac{1}{\sqrt{1 + \frac{1}{x^{2}}} + \sqrt{\frac{1}{x^{2}}}}\right).\) Multiply top and bottom of that fraction by $x$, so that it becomes \(\displaystyle \lim_{x\to0}\left( \frac{x}{\sqrt{x^2 + 1} + 1}\right).\) In that fraction, the numerator goes to $0$ as $x\to0$, but the denominator goes to $2$. So the fraction has limit $\dfrac02 = 0.$I'm guessing you can't do it simply this way? image001_zps75e88aa9.jpg Photo by phrox1 | Photobucket

- Thread starter
- #10

- Aug 30, 2012

- 1,199

I really should have written it asYou can not remove the limit because it is still indeterminate form . You can not simply cancel .

\(\displaystyle \lim_{x \to 0} \left ( \sqrt{ \frac{1}{x^2} } - \sqrt{ \frac{1}{x^2} } \right )\)

but I

-Dan