6
$\begingroup$

How to find the limit of $\frac{\sqrt{x^2+9}-3}{x^2}$? as x approaches 0

I thought that this was going to be a simple problem but then it got more complicated than I have expected

So I first multiplied the expression with its conjugate on both the denominator and the numerator the conjugate was the following:

$$\sqrt{x^2+9}+3$$

but when I multiplied by the conjugate I actually didn't even get close to the answer because the answer was supposed to be $\frac{1}{6}$

instead I have gotten the following by multiplying by its conjugate

$\frac{x^2+9-9}{x^2\sqrt{x^2+9}-3x^2}$ which obviously doesn't return 1/6

I was wondering if I was just simply making a mistake or if I have to choose a different option in solving this?

$\endgroup$
  • 7
    $\begingroup$ $\sqrt{x^2-9}$ is undefined when $x$ is near zero. It might make more sense if it was $\frac{\sqrt{9-x^2}-3}{x^2}$. $\endgroup$ – Thomas Andrews Jun 23 '17 at 2:28
  • 1
    $\begingroup$ One must suspect a typo, in which $\sqrt{9-x^2}$ was intended. $\endgroup$ – Michael Hardy Jun 23 '17 at 2:30
  • 1
    $\begingroup$ @ThomasAndrews & Michael Hardy I typed the problem wrong $\endgroup$ – John Rawls Jun 23 '17 at 2:33
  • $\begingroup$ I thought so. Even with the suggestion from both ThomasAndrews and MichaelHardy the answer would have been $-1/6$ i.e. different to what you have. Now the answer is $1/6$. $\endgroup$ – Rumplestillskin Jun 23 '17 at 2:35
  • 2
    $\begingroup$ The last term in the denominator should be $+3x^2$, not $-3x^2$. $\endgroup$ – mweiss Jun 23 '17 at 2:42
7
$\begingroup$

Perhaps you made a slight mistake when jumping from one step to the next. I have hidden the work, and tried my best to hide the solution. For some reason, you can't hide a block of text and the indented $\LaTeX$. If you're stuck, just place your mouse over the yellow tabs and the work should show up.


Multiplying by the conjugate is the right thing to do! Let's multiply the fraction by $\sqrt{x^2+9}+3$. Therefore

$$\lim\limits_{x\to0}\frac {\sqrt{x^2+9}-3}{x^2}=\lim\limits_{x\to0}\frac {(\sqrt{x^2+9}-3)(\sqrt{x^2+9}+3)}{x^2(\sqrt{x^2+9}+3)}=\lim\limits_{x\to0}\frac {x^2+9-9}{x^2(\sqrt{x^2+9}+3)}$$

The numerator reduces down to $x^2$, and we see that both $x^2$ terms cancel each other out. Can you finish the rest?

$$\lim\limits_{x\to0}\frac {1}{\sqrt{x^2+9}+3}=\frac 1{\sqrt9+3}=\frac 16$$


Edit: The OP has provided his work on the problem. The denominator is actually supposed to be $3x^2$ and not $-3x^2$ because we're multiplying by the conjugate: $\sqrt{x^2+9}+3$.

$\endgroup$
3
$\begingroup$

Note: \begin{align*} \frac{\sqrt{x^2+9}-3}{x^2} = & \; \frac{\big(\sqrt{x^2+9}-3 \big) \big(\sqrt{x^2+9}+3 \big)}{x^2 \big( \sqrt{x^2+9}+3 \big)} \\ = & \; \frac{x^2+9-3^2}{x^2\big(\sqrt{x^2+9}+3 \big)} \\ = & \; \frac{1}{\sqrt{x^2+9} + 3}. \end{align*} So we have \begin{align*} \lim_{x \to 0} \frac{\sqrt{x^2+9}-3}{x^2} = \lim_{x \to 0} \frac{1}{\sqrt{x^2+9} + 3} = \frac{1}{\sqrt{9}+3}=\frac{1}{6}. \end{align*} I guess you made some mistake when you multiplied the conjugate.

$\endgroup$
3
$\begingroup$

With $f (x)=\sqrt {x+9} $, the limit is

$$\lim_0\frac {f (x)-f (0)}{x}=f'(0) $$

and $$f'(x)=\frac {1}{2f (x)} $$

thus $$f'(0)=1/6$$

$\endgroup$
2
$\begingroup$

Let $f(t) = \sqrt{t}$. \begin{align*} \text{Then}\;\;f'(9)&=\lim_{h \to 0}\frac{\sqrt{9+h}-\sqrt{9}}{h}\\[4pt] &=\lim_{h \to 0^{+}}\frac{\sqrt{9+h}-\sqrt{9}}{h}\\[4pt] &=\lim_{x \to 0}\frac{\sqrt{9+x^2}-3}{x^2}\\[4pt] \end{align*}

$\endgroup$
  • 1
    $\begingroup$ Aren't you supposed to learn how to take limits before you learn how to take the derivative of certain functions? I believe that's how most schools teach people about integrals and derivatives. $\endgroup$ – Crescendo Jun 23 '17 at 2:49
  • 1
    $\begingroup$ @Crescendo I was the same. $\endgroup$ – Rumplestillskin Jun 23 '17 at 2:52
  • 1
    $\begingroup$ Usually limits are introduced first. After all, the standard definition of derivative is a limit. But after derivatives are introduced, as well as laws such as the power rule, some limits can be interpreted as derivatives, and then evaluated with very little work. $\endgroup$ – quasi Jun 23 '17 at 2:55
  • 1
    $\begingroup$ @Anonymous I think I read somewhere on MSE in the comments section that you don't want to blindly apply L'Hopitals. There could be much easier or standard ways of solving the limits. Even though L'Hopitals is a guaranteed success, I think. $\endgroup$ – Crescendo Jun 23 '17 at 3:16
  • 1
    $\begingroup$ but it's really straightforward $\frac{\sqrt{x^2+9}-3}{x^2} \rightarrow \frac{\frac{2x}{2\sqrt{x^2+9}}}{2x} = \frac{1}{2\cdot 3} = \frac{1}{6}$ $\endgroup$ – Anonymous Jun 23 '17 at 3:20
2
$\begingroup$

Since both the numerator and denominator go to 0, it is undefined. I would suggest using L'hospitals rule, which states that if there is an indeterminate form $\frac{0}{0}$ you can differentiate the numerator and the denominator and take the limit of the result. Here we have $\frac{\sqrt{x^2+9}-3}{x^2}$. Once we differentiate the numerator and denominator and simplify we obtain $\frac{1}{2\sqrt{x^2+9}}$. If we take the limit as $x$ goes to $0$, we get that the limit is equal to $\frac{1}{6}$.

$\endgroup$
  • $\begingroup$ Put \$ around your math symbols to make them appear typeset as equations. Such as \$\frac{a}{b}\$. $\endgroup$ – DanielV Jun 23 '17 at 2:52
  • $\begingroup$ @DanielV Curious, how to you get the dollar signs to show up without activating $\LaTeX$? $\endgroup$ – Crescendo Jun 23 '17 at 2:53
1
$\begingroup$

$$ \lim_{x\to0}\frac{\sqrt{x^2+9}-3}{x^2} = \lim_{x\to0}\frac{\frac{2x}{2\sqrt{x^2+9}}}{2x} = \frac{1}{2\cdot 3} = \frac{1}{6} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.