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Hi can anyone help me with this limit.

1) $\sqrt{5-\left(\frac{1}{\sqrt{1+\frac{y^2}{2}}}\right)}$ as $y\rightarrow -\infty $

I am struggling to do the first one, if it can be done using software then Maple, Mathematica or Matlab would be fine. I have tried l'hopitals rule but am just getting nowhere.

I would appreciate any help or suggestions.

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How do you define $\int_{x_0}^{x} \sqrt{f(x)}dx$? –  Did Feb 22 '13 at 8:38
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1 Answer

1) $$ \sqrt{1+\frac{y^2}{x^2}} \sim \sqrt{\frac{y^2}{x^2}} = -\frac y{|x|} \ (\text{when }y \rightarrow -\infty)\\ \sqrt{1-\frac y{x \sqrt{1+\frac {y^2}{x^2}}}} \sim \sqrt{1+\frac y{x \frac y{|x|}}} = \sqrt{1+\text{sign}(x)} = \left \{ \begin{array}{cc} \sqrt 2 & \text{if x > 0} \\ 0 & \text{if x < 0} \end{array}\right . $$ 2) I don't see any $y$ in your expression you're taking limit of.

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If it's plus sign then $$ \sqrt{1+\frac y{x \sqrt{1+\frac {y^2}{x^2}}}} \sim \sqrt{1-\text{sign}(x)} = \left \{ \begin{array}{cc} \sqrt 2 & \text{if x < 0} \\ 0 & \text{if x > 0} \end{array}\right . $$ –  Kaster Feb 22 '13 at 0:24
    
Also, would you edit your first post so $x \rightarrow \pm \infty$ is there. As for the limit itself, unfortunately "just a function" doesn't allow to take that limit. Different functions behave differently on $x \rightarrow \pm \infty$ –  Kaster Feb 22 '13 at 0:27
    
if $y \rightarrow +\infty$, then the only difference is $\sqrt{1+\frac{y^2}{x^2}} \sim \frac {y}{|x|}$. I'll leave rest of the calculations to you, since nothing is really changes, except sign. Can you also specify, what functions is under the integral - $\left ( a^2+a^2 \frac {y^2}{x^2}\right )^\frac 12$ or $\left ( a^2+a^2 \frac {y^2}{x^2}\right )^\frac 14$ ? –  Kaster Feb 22 '13 at 9:01
    
In this case $f(x) = a^2+a^2\frac{y^2}{x^2}$ not $f(x) = \sqrt{a^2+a^2\frac{y^2}{x^2}}$. –  Kaster Feb 22 '13 at 9:08
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