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Odd limit giving me troubles

phrox

New member
Sep 7, 2013
20

phrox

New member
Sep 7, 2013
20
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

Well-known member
MHB Math Helper
Aug 30, 2012
1,140
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 [tex]\infty - \infty[/tex]?

-Dan
 

phrox

New member
Sep 7, 2013
20
no I haven't covered L'Hospital's rule, also I can't directly sub x->0 into this because it just wont work, undefined.
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,140
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:
\(\displaystyle \frac{1}{x^2} + 1 \to \frac{1}{x^2}\)

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
 

chisigma

Well-known member
Feb 13, 2012
1,704
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

$\chi$ $\sigma$
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
\(\displaystyle \sqrt{\frac{1}{x^2}} - \sqrt{\frac{1}{x^2}} = 0\)

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

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
I'm guessing you can't do it simply this way? image001_zps75e88aa9.jpg Photo by phrox1 | Photobucket
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 $\dfrac02 = 0.$
 

phrox

New member
Sep 7, 2013
20
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

Well-known member
MHB Math Helper
Aug 30, 2012
1,140
You can not remove the limit because it is still indeterminate form . You can not simply cancel .
I really should have written it as
\(\displaystyle \lim_{x \to 0} \left ( \sqrt{ \frac{1}{x^2} } - \sqrt{ \frac{1}{x^2} } \right )\)

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

-Dan