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Find the limit $\lim_{x\to-1}\left(\frac{2x^2-x-3}{x^3+2x^2+6x+5}\right)$

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Final Answer

$-1$
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Step-by-step Solution

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1

We can factor the polynomial $x^3+2x^2+6x+5$ using the rational root theorem, which guarantees that for a polynomial of the form $a_nx^n+a_{n-1}x^{n-1}+\dots+a_0$ there is a rational root of the form $\pm\frac{p}{q}$, where $p$ belongs to the divisors of the constant term $a_0$, and $q$ belongs to the divisors of the leading coefficient $a_n$. List all divisors $p$ of the constant term $a_0$, which equals $5$

$1, 5$
2

Next, list all divisors of the leading coefficient $a_n$, which equals $1$

$1$
3

The possible roots $\pm\frac{p}{q}$ of the polynomial $x^3+2x^2+6x+5$ will then be

$\pm1,\:\pm5$
4

Trying all possible roots, we found that $-1$ is a root of the polynomial. When we evaluate it in the polynomial, it gives us $0$ as a result

${\left(-1\right)}^3+2\cdot {\left(-1\right)}^2+6\cdot -1+5=0$
5

Now, divide the polynomial by the root we found $\left(x+1\right)$ using synthetic division (Ruffini's rule). First, write the coefficients of the terms of the numerator in descending order. Then, take the first coefficient $1$ and multiply by the factor $-1$. Add the result to the second coefficient and then multiply this by $-1$ and so on

$\left|\begin{array}{c}1 & 2 & 6 & 5 \\ & -1 & -1 & -5 \\ 1 & 1 & 5 & 0\end{array}\right|-1$
6

In the last row of the division appear the new coefficients, with remainder equals zero. Now, rewrite the polynomial (a degree less) with the new coefficients, and multiplied by the factor $\left(x+1\right)$

$\lim_{x\to-1}\left(\frac{2x^2-x-3}{\left(x^{2}+x+5\right)\left(x+1\right)}\right)$
7

Factor the trinomial $2x^2-x-3$ of the form $ax^2+bx+c$, first, make the product of $2$ and $-3$

$\left(2\right)\left(-3\right)=-6$
8

Now, find two numbers that multiplied give us $-6$ and add up to $-1$

$\begin{matrix}\left(2\right)\left(-3\right)=-6\\ \left(2\right)+\left(-3\right)=-1\end{matrix}$
9

Rewrite the original expression

$\lim_{x\to-1}\left(\frac{2x^2-3x+2x-3}{\left(x^{2}+x+5\right)\left(x+1\right)}\right)$
10

Factor $2x^2-3x+2x-3$ by the greatest common divisor $2$

$\lim_{x\to-1}\left(\frac{2\left(x^2+x\right)-3x-3}{\left(x^{2}+x+5\right)\left(x+1\right)}\right)$
11

Factor $2\left(x^2+x\right)-3x-3$ by the greatest common divisor $3$

$\lim_{x\to-1}\left(\frac{2\left(x^2+x\right)-3\left(x+1\right)}{\left(x^{2}+x+5\right)\left(x+1\right)}\right)$
12

Factor the polynomial $\left(x^2+x\right)$ by it's greatest common factor (GCF): $x$

$\lim_{x\to-1}\left(\frac{2x\left(x+1\right)-3\left(x+1\right)}{\left(x^{2}+x+5\right)\left(x+1\right)}\right)$
13

Factor the polynomial $2x\left(x+1\right)-3\left(x+1\right)$ by it's greatest common factor (GCF): $x+1$

$\lim_{x\to-1}\left(\frac{\left(x+1\right)\left(2x-3\right)}{\left(x^{2}+x+5\right)\left(x+1\right)}\right)$
14

Simplify the fraction $\frac{\left(x+1\right)\left(2x-3\right)}{\left(x^{2}+x+5\right)\left(x+1\right)}$ by $x+1$

$\lim_{x\to-1}\left(\frac{2x-3}{x^{2}+x+5}\right)$
15

Evaluate the limit $\lim_{x\to-1}\left(\frac{2x-3}{x^{2}+x+5}\right)$ by replacing all occurrences of $x$ by $-1$

$\frac{-2-3}{{\left(-1\right)}^{2}-1+5}$
16

Subtract the values $5$ and $-1$

$\frac{-2-3}{4+{\left(-1\right)}^{2}}$
17

Subtract the values $-2$ and $-3$

$\frac{-5}{4+{\left(-1\right)}^{2}}$
18

Calculate the power ${\left(-1\right)}^{2}$

$\frac{-5}{4+1}$
19

Add the values $4$ and $1$

$-\frac{5}{5}$
20

Divide $-5$ by $5$

$-1$

Final Answer

$-1$

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Limits by Direct SubstitutionLimits by L'Hôpital's ruleLimits by FactoringLimits by Rationalizing

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a
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g
m
n
u
v
w
x
y
z
.
(◻)
+
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×
◻/◻
/
÷
2

e
π
ln
log
log
lim
d/dx
Dx
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

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