Different Way to Solve a Limit of an Indeterminate Equation

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<br /> lim_{x\rightarrow-2}\frac{x^3 + 8}{x+2}<br />

Ok, I have solved this using synthetic division in which the limit was 12, however I was wondering if there was a way to solve this without using synthetic division (seeing as I hate the process of it).

I've solved plenty of other indeterminate equations by factoring out a polynomial and then being able to cancel out so that I can then plug in x, but I wasn't able to do so with this one.

I'm just curious if there are any other ways to solve this specific limit.

Thanks
 
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L'hospital rule (applies only when it is in forms like 0/0, inf/inf)

differentiate both num and den, and then sub in your -2
 
Yeah I started reading up on that rule, but didn't understand it fully until your reply. So the derivative of the numerator is 3x^2 and the denominator is 1 so then just sub in the -2 and you get 12. Got it, thanks a lot for clearing up that rule for me.
 
Not that this is much different than using synthetic division, but the numerator is a sum of two cubes, which factors according to

a^3 + b^3 = (a + b) \cdot (a^2 - ab + b^2) . So, in fact, this numerator can be factored and the (x + 2) term can be cancelled.
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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