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$\lim_{x\to 2} {x^2 - 4\over x^3 - 4x^2 +4x}$

I used L'Hospital's rule twice on this, and got a solution, but my textbook says it's an indeterminate form. Is using L'Hospital's rule twice wrong, and if yes, why so?

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Edited answer after the correction of the OP :

We have the limit :

$$\lim_{x \to 2} \frac{x^2-4}{x^3-4x^2 + 4x} $$

Note that this is an indeterminate form, thus L'Hospital's can be applied :

$$\lim_{x \to 2} \frac{x^2-4}{x^3-4x^2 + 4x} = \lim_{x \to 2} \frac{2x}{3x^2-8x + 4}$$

Note that this is not an indeterminate form, so D.L.H. cannot be applied.

Now, you have to decide whether this limit exists.

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  • $\begingroup$ Sorry everyone, I missed an x in the question. My bad. $\endgroup$ – Green05 Mar 4 at 17:00
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You can use L'Hoptital rules as many times as you like so long as the numerator and denominator make an inderterminate form.

$\frac {x^2 -4}{x^3-4x^2 +4x}_{x=2}\to \frac {2^2-4}{2^3 - 4*2^2 + 4*2} \to \frac 00$.

So we can use it.

First time:

$\frac {(x^2 -4)'}{(x^3 - 4x^2 + 4x)}=\frac {2x}{3x^2 -8x+4}_{x=2} \to \frac {4}{3*4-8*2+ 4} = \frac {4}{0}$.

That is not indeterminate form because the numerator is $4$.

So you can't use it twice.

....

On the other hand if had $\lim\limits_{x\to 2}\frac {x^3 -6x^2 + 12x - 8}{x^4 - 8x^3 + 24x^2 -32x + 16}$ you could use L'Hopital $3$ times.

$\frac {x^3 -6x^2 + 12x - 8}{x^4 - 8x^3 + 24x^2 -32x + 16}_{x=2}\to\frac 00$

$\frac {3x^2 -12x + 12}{4x^3-24x^2 + 48x -32}_{x=2}\to \frac 00$

$\frac{6x-12}{12x^2 - 48x + 48}_{x=0}\to \frac 00$

$\frac 6{24x - 48}_{x=2} \to \frac 60$.

So $\lim\limits_{x\to 2}\frac {x^3 -6x^2 + 12x - 8}{x^4 - 8x^3 + 24x^2 -32x + 16}=\infty$.

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L'hopital rule only applies if the limit is indeterminate.

Here it is simply 0/(-4) = 0.

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