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

Step-by-step Solution

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Final Answer

$2$
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Step-by-step Solution

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1

Apply the power rule for limits: $\lim_{x\to a}\left(f(x)\right)^n=\left(\lim_{x\to a}f(x)\right)^n$

$\sqrt{\lim_{x\to\infty }\left(\frac{x+8x^2}{2x^2-1}\right)}$

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$\sqrt{\lim_{x\to\infty }\left(\frac{x+8x^2}{2x^2-1}\right)}$

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Learn how to solve problems step by step online. Find the limit of ((x+8x^2)/(2x^2-1))^1/2 as x approaches infinity. Apply the power rule for limits: \lim_{x\to a}\left(f(x)\right)^n=\left(\lim_{x\to a}f(x)\right)^n. If we directly evaluate the limit \lim_{x\to \infty }\left(\frac{x+8x^2}{2x^2-1}\right) as x tends to \infty , we can see that it gives us an indeterminate form. We can solve this limit by applying L'H么pital's rule, which consists of calculating the derivative of both the numerator and the denominator separately. After deriving both the numerator and denominator, the limit results in.

Final Answer

$2$

Exact Numeric Answer

$2$

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Limits by Direct SubstitutionLimits by L'H么pital's ruleLimits by FactoringLimits by Rationalizing

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1
2
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

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