Find the limit of $\frac{2y^2-3y+5}{y^2-5y+2}$ as $y$ approaches $\infty $

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Solving: $\lim_{y\to\infty }\left(\frac{2y^2-3y+5}{y^2-5y+2}\right)$

Final answer to the problem

$2$
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Step-by-step Solution

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1

As it's an indeterminate limit of type $\frac{\infty}{\infty}$, divide both numerator and denominator by the term of the denominator that tends more quickly to infinity (the term that, evaluated at a large value, approaches infinity faster). In this case, that term is

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$\lim_{y\to\infty }\left(\frac{\frac{2y^2-3y+5}{y^2}}{\frac{y^2-5y+2}{y^2}}\right)$

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Learn how to solve limits to infinity problems step by step online. Find the limit of (2y^2-3y+5)/(y^2-5y+2) as y approaches infinity. As it's an indeterminate limit of type \frac{\infty}{\infty}, divide both numerator and denominator by the term of the denominator that tends more quickly to infinity (the term that, evaluated at a large value, approaches infinity faster). In this case, that term is . Separate the terms of both fractions. Simplify the fraction . Simplify the fraction by y.

Final answer to the problem

$2$

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Function Plot

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

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3
4
5
6
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 to Infinity

The limit of a function f(x) when x tends to infinity is the value that the function takes as the value of x grows indefinitely.

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