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Again, I have a problem in the indeterminate form and I can't find out how to reduce this to something that'll give me the limit.

$$\lim_{ x\to-2} \frac{x^2-x-6}{x^2+x-2}=\frac{(-2)^2+2-6}{(-2)^2-2-2}=\frac{4+2-6}{4-2-2}=\frac{0}{0}$$

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Do you know how to factor (quadratic) polynomials? If it makes it easier to factor, $x+2$ must divide both the numerator and the denominator. (Do you see why?) –  Srivatsan Sep 2 '11 at 3:52
    
Since the polynomials evaluate to $0$ at $x=-2$, they can both be factored as $(x+2)(\text{something})$. Find the something for each, cancel and take the limit again. –  Arturo Magidin Sep 2 '11 at 3:52

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HINT $\ $ Consider the limit of a rational function $\rm\:f(x)/g(x)\:$ as $\rm\:x\to c\:.\:$ If $\rm\:f(c)=0=g(c)\:$ then, by the Factor Theorem, both $\rm\:f,\: g\:$ are divisible by $\rm\:x-c\:.\:$ Keep cancelling $\rm\:x-c\:$ from both $\rm\:f,\: g\:$ till this is no longer possible. Then at least one of $\rm\:f,\: g\:$ will no longer be divisible by $\rm\:x-c\:,\:$ so they will no longer both have $\rm\:c\:$ as a root, hence the limit will no longer be of indeterminate form $\:0/0\:.\:$ Thus you can always remove such "apparent singularities" from rational functions by simply cancelling such common factors.

Applying this to your specific example we deduce that it has numerator and denominator both divisible by $\rm\:x + 2\:.\:$ What happens when you proceed as above and cancel this common factor?

Later you'll learn about l'Hôpital's rule - which provides a more general method to resolve such indeterminate forms. For the special case of rational functions, l'Hospital's rule is essentially equivalent to the cancellation method described above.

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