1. Compute the following limits using L'Hôpital's rule.
(a) $\lim _{x \rightarrow 2} \frac{x^{3}-3 x^{2}+x+2}{x^{2}+2 x-8}$
(b)

Answer

Expert–verified

Hide Steps

Answer

$\boxed{\lim_{x \rightarrow 2} \frac{x^3 - 3x^2 + x + 2}{x^2 + 2x - 8} = \frac{1}{6}}$

Steps

Step 1 ：Compute the derivatives of the functions in the numerator and denominator: $f'(x) = 3x^2 - 6x + 1$ and $g'(x) = 2x + 2$

Step 2 ：Apply L'Hôpital's rule: $\lim_{x \rightarrow 2} \frac{f'(x)}{g'(x)} = \lim_{x \rightarrow 2} \frac{3x^2 - 6x + 1}{2x + 2}$

Step 3 ：Calculate the limit: $\lim_{x \rightarrow 2} \frac{3x^2 - 6x + 1}{2x + 2} = \frac{1}{6}$

Step 4 ：$\boxed{\lim_{x \rightarrow 2} \frac{x^3 - 3x^2 + x + 2}{x^2 + 2x - 8} = \frac{1}{6}}$