Find the Limit of (1/(x-2) - 12/(x^3-8)) as x Approaches 2 - Homework Help

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Homework Statement


find the limit of:

\lim_{x\rightarrow2}\frac{1}{x-2}-\frac{12}{x^{3}-8}




The Attempt at a Solution



Any help guys? Should i cross multiply then divide and see what i get or is there something else.(besides l'hospital's rule and taylor series because i haven't learned them yet.)
 
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I can't imagine why you would think "cross multiply" here. Of course, what you should do is the subtraction indicated in the problem"
\frac{1}{x- 2}- \frac{12}{x^3- 8}= \frac{?}{x^3- 8}
 
mtayab1994 said:

Homework Statement


find the limit of:

\lim_{x\rightarrow2}\frac{1}{x-2}-\frac{12}{x^{3}-8}




The Attempt at a Solution



Any help guys? Should i cross multiply then divide and see what i get or is there something else.
Cross multiply? You do that when you have an equation that involves two fractions.

Combine the two fractions, and then go from there. Note that the difference of cubes can be factored: a3 - b3 = (a - b)(a2 + ab + b2).
mtayab1994 said:
(besides l'hospital's rule and taylor series because i haven't learned them yet.)
L'Hopital's Rule cannot be applied to this problem because what you have is a difference, not a quotient.
 
ok i got:

\lim_{x\rightarrow2}\frac{x+4}{x^{2}+2x+4}=\frac{1}{2}

is that correct?
 
Thanx.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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