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I'm making a bunch of limits exercises and I found me stuck with this limit:

$$\begin{align} \lim_{x\to 8} {\frac{64-x^2}{8-x}} \end{align}$$

It gives me an indetermination of

$$\begin{align} \frac{0}{0} \end{align}$$

So, the next step is to factorize. I used the Ruffini's Rule.

The solution is $16$. And my result was $-16$, something is wrong :(

Any help?

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  • $\begingroup$ Use $a^2-b^2=(a-b)(a+b)$. $\endgroup$ – Git Gud Mar 25 '14 at 23:27
  • $\begingroup$ It should be clear that if you approach from one side both the numerator and the denominator will have the same sign so the limit (if it exists, which it does) will be positive. $\endgroup$ – Brad Mar 25 '14 at 23:33
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To use Ruffini's Rule, you need to use x-8, not 8-x. So, multiply both numerator and denominators by -1 one and then apply Ruffini's rule. You will get the correct answer, which is +16. Another trick for futur reference, is to compute the derivatives of both numerators and denominators and then compute the limit. In this case, you will get -2x divided by -1, which gives you 2x. When you replace x with 8, you get the answer +16.

Note: I saw that you said Ruffini's rule gives you -x-8, I think that is where your error is: when you multiply by -1, you should get -(8-x) = x-8. Same applies to the numerator.

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$$\frac{64-x^2}{8-x}=\frac{(8-x)(8+x)}{8-x}=8+x \to 16$$ Did you understand?

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  • $\begingroup$ Yes, but what about Ruffini's rule? It gives me (-x-8) $\endgroup$ – Nuno Batalha Mar 25 '14 at 23:31
  • $\begingroup$ Ruffini's rule requires division by something that looks like $x-a$. Here you're dividing by $a-x$ so you need to make adjustments. $\endgroup$ – Git Gud Mar 25 '14 at 23:33
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$$\begin{align} \lim_{x\to 8} {\frac{64-x^2}{8-x}} \end{align}$$

Well, isn't THIS an easy limit! Obviously, we can't just plug it in. We'll get $\frac{0}{0}$! Oh no... So what else can we do? Hmm... $x^2$... $64$ which is kind of relevant to $8$... Actually, WAIT, YOU CAN FACTOR!

$$64-x^2=(8-x)(8+x)$$

This is a simple difference of squares factoring. And what do you know, we have $8-x$ there just like in the denominator! Let's plug it back in...

$$\lim_{x\to 8} {\frac{64-x^2}{8-x}}=\lim_{x\to 8} {\frac{(8-x)(8+x)}{8-x}}=\lim_{x\to 8} (8+x)$$

YES! Now we don't have any problems! We can just plug $x$ in!!! THIS IS AN EXCITING MOMENT:

$$\lim_{x\to 8} (8+x)=(8+8)=16$$

Another rule you should know is L'Hospital's Rule, which involves a little differentiation but may be useful:

$$if \lim_{x\to \zeta}\frac{f(x)}{g(x)}=\frac{0}{0} \text{ or } \frac{\infty}{\infty}, \lim_{x\to \zeta}\frac{f(x)}{g(x)}=\lim_{x\to \zeta}\frac{f'(x)}{g'(x)}$$

Basically, if you get $\frac{0}{0} \text{ or } \frac{\infty}{\infty}$ when you plug in, and both the numerator and denominator are differential (i.e. contain variable), then you can take the derivative of the top and that of the bottom and get the limit from there.

Thus:

$$\lim_{x\to 8} {\frac{64-x^2}{8-x}} =\lim_{x\to 8} {\frac{0-2x}{0-1}}=\lim_{x\to 8} {2x}=16 $$

Incredible. Amazing. Stunning. Sophisticated at high levels. So much better than Lagrange.

CHEERS! -Shahar

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When you used Ruffini's Rule, did you remember to divide by the leading coefficient when you are done?

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  • $\begingroup$ Yes, in the numerator it gave me (-x-8) $\endgroup$ – Nuno Batalha Mar 25 '14 at 23:34
  • $\begingroup$ If you use Ruffini's Rule, you should have ended up with x+8, not -x-8. You missed a negative somewhere. First , we synthetically divide by the root of the denominator and get -x-8. But then we have to divide by the leading coefficient of negative one. So, the final answer is x+8. $\endgroup$ – Bobby Ocean Mar 25 '14 at 23:42
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$$ \frac{64 - x^2}{8-x} = 8 + x \;, $$ so, the limit as $x \rightarrow 8$ is of course $16$.

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