While helping a friend out for an exam (last year of high school), I found an exercise that neither of us could solve. I've tried a couple of different approaches but nothing seemed to work. Could anyone tell me how to solve it, or at least some hints? This is the exercise:

Knowing that

$$\lim_{x \to a}\frac{x^2-\sqrt{a^3x}}{\sqrt{ax}-a}=12$$

Find out $a$.

We can assume that $a$ exists and is real.


Besides L'Hopital one can simply rationalize the denominator, after discarding case $\rm\:a \le 0\:,\:$ viz:

$$\rm \frac{x^2-a\sqrt{ax}}{\sqrt{ax}-a}\ =\ \frac{x^2-a\sqrt{ax}}{\sqrt{ax}-a} \ \frac{\sqrt{ax}+a}{\sqrt{ax}+a}\ =\ \frac{ax\:(x-a)+\sqrt{ax}\ (x^2-a^2) }{a\:(x-a) }\ =\ x+(x+a)\sqrt{\frac{x}{a}}$$

Since the above $\rm\to 3\ a\ $ as $\rm\ x\to a\ $ the problem reduces to solving $\rm\ 3\ a = 12\:$.

| cite | improve this answer | |
  • $\begingroup$ Now that I see it, this is definitely the way we were taught to do this (it was quite some time ago, I couldn't remember). One thing I don't fully understand, though, is why you assume a isn't negative. $\endgroup$ – Javier Feb 27 '11 at 1:23
  • $\begingroup$ @Jav: See Sivaram's answer for $\rm a\le 0$. $\endgroup$ – Bill Dubuque Feb 27 '11 at 1:26
  • $\begingroup$ So it means that for any $a < 0$ the limit will always be 0 and therefore not relevant (because it's not 12)? $\endgroup$ – Javier Feb 27 '11 at 3:11
  • $\begingroup$ @Jav: Yes, that's it. $\endgroup$ – Bill Dubuque Feb 27 '11 at 3:32

First note that $a>0$.

If $a<0$, then $$\lim_{x \rightarrow a} \frac{x^2-\sqrt{a^3x}}{\sqrt{ax}-a} = \frac{a^2 - a^2}{-a-a} = 0$$

If $a=0$, then $\lim_{x \rightarrow a} \frac{x^2-\sqrt{a^3x}}{\sqrt{ax}-a}$ doesn't exist.

Hence, $a>0$.

Let $\sqrt{ax}=y$ i.e. $x = \frac{y^2}{a}$. Note that as $x \rightarrow a$, $y \rightarrow a$.

Hence, $$\lim_{x \rightarrow a} \frac{x^2-\sqrt{a^3x}}{\sqrt{ax}-a} = \lim_{y \rightarrow a} \frac{y}{a^2} \frac{y^3-a^3}{y-a} = \frac{3a^2}{a} = 3a$$

Hence, $a=4$

| cite | improve this answer | |
  • $\begingroup$ This seems to be the way I was meant to solve it. If no answer comes up in a while, I'll accept this one. Thank you. $\endgroup$ – Javier Feb 26 '11 at 22:41
  • 1
    $\begingroup$ This is a nice approach. It is clean, and avoids technical manipulations (such as using L'Hôpital's rule). $\endgroup$ – Andrés E. Caicedo Feb 26 '11 at 23:40

One way is application of L'hoipitals rule. Since both denominator and numerator in $$\lim_{x \to a}\frac{x^2-\sqrt{a^3x}}{\sqrt{ax}-a}$$ go to zero, by l'hopitals rule this is the same as $$\lim_{x \to a}\frac{2x-a^{3/2}\frac{1}{2}x^{-1/2}}{\frac{\sqrt{a}}{2}x^{-1/2}}$$ Simplifying the expression we get that this equals

$$\lim_{x \to a}\frac{4x^{3/2}-a^{3/2}}{\sqrt{a}}=\frac{3a^{3/2}}{\sqrt{a}}=3a$$

Thus choose $a=4$.

| cite | improve this answer | |
  • $\begingroup$ Oh, I didn't know about L'Hôpital's rule. Thank you, but I don't think this is how I'm supposed to solve it. Thanks anyway. $\endgroup$ – Javier Feb 26 '11 at 22:31

Using L'hopital Rule we have $$ \lim_{x \to a} \frac{2x- \frac{a^3}{2 \sqrt{a^{3}x}}}{\frac{a}{2 \sqrt{ax}}} = 12$$

$$\Longrightarrow \frac{2a-\frac{a^3}{2a^2}}{\frac{1}{2}} = 12$$

$$\Longrightarrow 2\left(2a- \frac{a^3}{2a^2}\right)= 12$$

So $a = 4$.

| cite | improve this answer | |
  • $\begingroup$ Could you explain how you got from the original formula to your first step? $\endgroup$ – Javier Feb 26 '11 at 22:28
  • $\begingroup$ Thanks, but the same thing I said to Eric Naslund applies. Thanks anyway. $\endgroup$ – Javier Feb 26 '11 at 22:39

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.