Find $$\lim_{x \to 2}\frac{x^3+3x^2-12x+4}{x^3-4x}$$ How do you get to your solution? Thanks in advance. I have tried factoring out $x^3$ to no avail.
Is there an algebraic solution? I don't know L'H rule.
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L'H rule is applicable. – Sandeep Silwal Apr 13 '14 at 02:03
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@Artes I'm not sure what you're referring to. Can you please clarify? – user140900 May 11 '14 at 02:02
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$$\lim_{x \to 2}\frac{x^3+3x^2-12x+4}{x^3-4x}=\lim_{x \to 2}\frac{(x-2)(x^2+5x-2)}{x(x+2)(x-2)}=3/2$$
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$$\lim_{x \to 2}\frac{x^3+3x^2-12x+4}{x^3-4x}=\lim_{x \to 2}\frac{3x^2+6x-12}{3x^2-4}=\dfrac{3}{2}$$ In the second step, I used L'Hopital's rule.