Final Answer
$-1$
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Step-by-step Solution
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1
If we directly evaluate the limit $\lim_{x\to -1}\left(\frac{2x^2-x-3}{x^3+2x^2+6x+5}\right)$ as $x$ tends to $-1$, we can see that it gives us an indeterminate form
$\frac{0}{0}$
2
We can solve this limit by applying L'H么pital's rule, which consists of calculating the derivative of both the numerator and the denominator separately
$\lim_{x\to -1}\left(\frac{\frac{d}{dx}\left(2x^2-x-3\right)}{\frac{d}{dx}\left(x^3+2x^2+6x+5\right)}\right)$
3
After deriving both the numerator and denominator, the limit results in
$\lim_{x\to-1}\left(\frac{4x-1}{3x^{2}+4x+6}\right)$
4
Evaluate the limit $\lim_{x\to-1}\left(\frac{4x-1}{3x^{2}+4x+6}\right)$ by replacing all occurrences of $x$ by $-1$
$\frac{-4-1}{3\cdot {\left(-1\right)}^{2}-4+6}$
5
Subtract the values $6$ and $-4$
$\frac{-4-1}{2+3\cdot {\left(-1\right)}^{2}}$
6
Subtract the values $-4$ and $-1$
$\frac{-5}{2+3\cdot {\left(-1\right)}^{2}}$
7
Calculate the power ${\left(-1\right)}^{2}$
$\frac{-5}{2+3}$
8
Add the values $2$ and $3$
$-\frac{5}{5}$
9
Divide $-5$ by $5$
$-1$
Final Answer
$-1$