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Find the limit of $\frac{x^2+6x+5}{x^2-2x-3}$ as $x$ approaches $-1$

Step-by-step Solution

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Final answer to the problem

$-1$
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Step-by-step Solution

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  • Solve using L'Hôpital's rule
  • Solve without using l'Hôpital
  • Solve using limit properties
  • Solve using direct substitution
  • Solve the limit using factorization
  • Solve the limit using rationalization
  • Integrate by partial fractions
  • Product of Binomials with Common Term
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1

Factor the trinomial $x^2-2x-3$ finding two numbers that multiply to form $-3$ and added form $-2$

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

Learn how to solve limits by direct substitution problems step by step online.

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

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Learn how to solve limits by direct substitution problems step by step online. Find the limit of (x^2+6x+5)/(x^2-2x+-3) as x approaches -1. Factor the trinomial x^2-2x-3 finding two numbers that multiply to form -3 and added form -2. Rewrite the polynomial as the product of two binomials consisting of the sum of the variable and the found values. Factor the trinomial x^2+6x+5 finding two numbers that multiply to form 5 and added form 6. Rewrite the polynomial as the product of two binomials consisting of the sum of the variable and the found values.

Final answer to the problem

$-1$

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Function Plot

Plotting: $\frac{x^2+6x+5}{x^2-2x-3}$

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5
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7
8
9
0
a
b
c
d
f
g
m
n
u
v
w
x
y
z
.
(◻)
+
-
×
◻/◻
/
÷
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

How to improve your answer:

Main Topic: Limits by Direct Substitution

Find limits of functions at a specific point by directly plugging the value into the function.

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