Factoring Limits: Solving Quadratic Equations

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    Factoring Limits
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Factoring limits in calculus often involves simplifying expressions to evaluate them at specific points. In the example provided, the limit of (x^2 - 5x - 6)/(x^2 - 4) can be factored to (x-3)(x-2) to find the limit as x approaches 2. For the second example, (x^2 + 3x - 24)/(x^2 + 8), direct substitution of x = 4 is possible since it does not result in an undefined expression, making factoring unnecessary. Understanding when to factor versus when to substitute is key in solving limit problems. Mastery of these techniques is essential for success in algebra and calculus.
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Okay so I've been teaching myself (with the aid of the mighty internet & several friends) algebra & now calculus. I have found that I didn't do too good at high school for various reasons. Some good...some not good. Anyway...

I have a (what is probably a basic question) about factoring limit questions. I understand that with an equation like below (which results in 0, you need to factor it)

Lim x^2 -5x - 6
x->2 ------------
x^2 - 4

factored it works out to (x-3)(x-2).

I can manage that. What is the approach taken when you have something like

Lim x^2 + 3x -24
x->4 ------------
x^2 + 8

I don't know what to do in a situation like this, I don't know how to break it into a (x a)(x b) situation. I hope this thread isn't looked at & thought 'what a twit' I should say that I am aware of the quadratic equation...
 
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In this problem you can actually just substitute x = 4 since the result is defined (no 0 in the denominator or an infinity anywhere) :)
 

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