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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.

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up vote 5 down vote accepted

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 $\rm\ 3\ a = 12\:$.

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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. – Javier Feb 27 '11 at 1:23
@Jav: See Sivaram's answer for $\rm a\le 0$. – Bill Dubuque Feb 27 '11 at 1:26
So it means that for any $a < 0$ the limit will always be 0 and therefore not relevant (because it's not 12)? – Javier Feb 27 '11 at 3:11
@Jav: Yes, that's it. – 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, $a=4$

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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. – Javier Feb 26 '11 at 22:41
This is a nice approach. It is clean, and avoids technical manipulations (such as using L'Hôpital's rule). – 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 $a=4$.

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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. – 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 $a = 4$.

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Could you explain how you got from the original formula to your first step? – Javier Feb 26 '11 at 22:28
Thanks, but the same thing I said to Eric Naslund applies. Thanks anyway. – Javier Feb 26 '11 at 22:39

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