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Find the limit of $\frac{x^2-7x+10}{x-5}$ as $x$ approaches $5$

Step-by-step Solution

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Solution

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Final answer to the problem

$3$
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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
  • FOIL Method
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1

Factor the trinomial $x^2-7x+10$ finding two numbers that multiply to form $10$ and added form $-7$

$\begin{matrix}\left(-2\right)\left(-5\right)=10\\ \left(-2\right)+\left(-5\right)=-7\end{matrix}$

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

$\begin{matrix}\left(-2\right)\left(-5\right)=10\\ \left(-2\right)+\left(-5\right)=-7\end{matrix}$

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Learn how to solve limits by direct substitution problems step by step online. Find the limit of (x^2-7x+10)/(x-5) as x approaches 5. Factor the trinomial x^2-7x+10 finding two numbers that multiply to form 10 and added form -7. Rewrite the polynomial as the product of two binomials consisting of the sum of the variable and the found values. Simplifying. Evaluate the limit \lim_{x\to5}\left(x-2\right) by replacing all occurrences of x by 5.

Final answer to the problem

$3$

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Function Plot

Plotting: $\frac{x^2-7x+10}{x-5}$

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0
a
b
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d
f
g
m
n
u
v
w
x
y
z
.
(◻)
+
-
×
◻/◻
/
÷
◻2
◻◻
√◻
√

◻√◻
◻√
∞
e
π
ln
log
log◻
lim
d/dx
D□x
∫
∫◻
|◻|
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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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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