Evaluate the limit
$\lim_{x\to0}\left(\frac{\sin\left(3x\right)}{\tan\left(4x\right)}\right)$
$factor\left(x^4+x^3-6x^2-4x+8\right)$
$\lim_{x\to0}\left(\frac{x^2}{1-\cos\left(x\right)}\right)$
$factor\left(121+198x^6+81x^{12}\right)$
$\lim_{x\to0}\left(\frac{\sqrt{5+x}-\sqrt{5}}{x}\right)$
$\lim_{x\to10}\left(\frac{\sqrt{x+6}-4}{x-10}\right)$
$\lim_{x\to1}\left(\frac{\sqrt{x+3}-2}{x^2-1}\right)$
Limits by factoring
~ 0.79 s
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