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I'm being asked to evaluate the limit of:

$$\lim_{x\to 4}\frac{x^2+5x+4}{x^2+3x-4}$$

There aren't any solutions, so I'm just wondering if I'm going about this properly, I first factored the numerator and denominator to:

$$\frac{(x+4)(x+1)}{(x+4)(x-1)}\;.$$

then cancelled both of the $(x+4)$ and plugged in $4$ for the $x$'s to get an answer of $5/3$.

Am I doing this right?

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1  
You are indeed. I wonder, however, whether the limit was actually as $x\to -4$, as then the factorization would actually be necessary. –  Brian M. Scott Feb 3 '13 at 0:04
    
Yes. This is correct. –  Inquest Feb 3 '13 at 0:05

3 Answers 3

up vote 4 down vote accepted

Your calculation is correct, but you may have missed something important. As $x$ approaches $4$, the unmodified denominator approaches a perfectly respectable non-zero number. So the factorization was a waste of time, and irrelevant.

If the question were about the limit as $x$ approaches $-4$, that would be another matter. The factorization and cancelling would be very useful.

Remark: Look at the denominator $x^2+3x-4$. Imagine $x$ approaching $4$. Then $x^2$ approaches $16$, and $3x$ approaches $12$, so $x^2+3x-4$ approaches $24$, a nice non-zero number.

Suppose that the numerator was instead something like $x^2+x+1$, which does not have $x+4$ as a factor. Then you might be puzzled about what to do. But there is a simple answer. As $x$ approaches $4$, $x^2+x+1$ approaches $21$, so $\frac{x^2+x+1}{x^2+3x-4}$ approaches $\frac{21}{24}$.

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$$\lim_{x\to 4}\frac{x^2+5x+4}{x^2+3x-4}=\lim_{x\to 4}\frac{x^2+x+4x+4}{x^2-x+4x-4}=$$ $$=\lim_{x\to 4}\frac{x(x+1)+4(x+1}{x(x-1)+4(x-1)}=\lim_{x\to 4}\frac{(x+1)(x+4)}{(x-1)(x+4)}=\lim_{x\to 4}\frac{(x+1)}{(x-1)}=\frac{5}{3}$$ Directly $$\lim_{x\to 4}\frac{x^2+5x+4}{x^2+3x-4}=\frac{40}{24}=\frac{5}{3}$$ and $$\lim_{x\to -4}\frac{x^2+5x+4}{x^2+3x-4}=\lim_{x\to -4}\frac{(x+1)}{(x-1)}=\frac{3}{5}$$

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$$\lim_{x\to a}\frac{f(x)}{g(x)}=\frac{\lim_{x\to a}f(x)}{\lim_{x\to a}g(x)}$$ if both $\lim_{x\to a}f(x)$ and $\lim_{x\to a}g(x)$ exists and $\lim_{x\to a}g(x)\neq0$

In your problem, both limits exists and denominator limit is non-zero, thus, you don't need to do any fancy thing like factorization,just plug in the value.

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