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Find the limit of $\sqrt{\frac{x+8x^2}{2x^2-1}}$ as $x$ approaches $\infty $

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Function Plot

Plotting: $\sqrt{\frac{x+8x^2}{2x^2-1}}$

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8
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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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