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Limit of the function/ x+sqrt(x)

Limit of the function x+sqrt(x)

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 lim \x + \/ x /
x->oo
lim⁡x→∞(x+x)\lim_{x \to \infty}\left(\sqrt{x} + x\right) 
Limit(x + sqrt(x), x, oo, dir='-')
Detail solution
Let's take the limit
lim⁡x→∞(x+x)\lim_{x \to \infty}\left(\sqrt{x} + x\right)
Let's eliminate indeterminateness oo - oo
Multiply and divide by
x−x\sqrt{x} - x
then
lim⁡x→∞(x+x)\lim_{x \to \infty}\left(\sqrt{x} + x\right)
=
lim⁡x→∞((x−x)(x+x)x−x)\lim_{x \to \infty}\left(\frac{\left(\sqrt{x} - x\right) \left(\sqrt{x} + x\right)}{\sqrt{x} - x}\right)
=
lim⁡x→∞((x)2−(−x)2x−x)\lim_{x \to \infty}\left(\frac{\left(\sqrt{x}\right)^{2} - \left(- x\right)^{2}}{\sqrt{x} - x}\right)
=
lim⁡x→∞(−x2+xx−x)\lim_{x \to \infty}\left(\frac{- x^{2} + x}{\sqrt{x} - x}\right)
=
lim⁡x→∞(−x2+xx−x)\lim_{x \to \infty}\left(\frac{- x^{2} + x}{\sqrt{x} - x}\right)

Let's divide numerator and denominator by sqrt(x):
lim⁡x→∞(−x32+x1−x)\lim_{x \to \infty}\left(\frac{- x^{\frac{3}{2}} + \sqrt{x}}{1 - \sqrt{x}}\right) =
lim⁡x→∞(−x32+x1−x)\lim_{x \to \infty}\left(\frac{- x^{\frac{3}{2}} + \sqrt{x}}{1 - \sqrt{x}}\right) =
lim⁡x→∞(−x32+x1−x)\lim_{x \to \infty}\left(\frac{- x^{\frac{3}{2}} + \sqrt{x}}{1 - \sqrt{x}}\right)
Do replacement
u=1xu = \frac{1}{x}
then
lim⁡x→∞(−x32+x1−x)\lim_{x \to \infty}\left(\frac{- x^{\frac{3}{2}} + \sqrt{x}}{1 - \sqrt{x}}\right) =
lim⁡u→0+(−(1u)32+1u1−1u)\lim_{u \to 0^+}\left(\frac{- \left(\frac{1}{u}\right)^{\frac{3}{2}} + \sqrt{\frac{1}{u}}}{1 - \sqrt{\frac{1}{u}}}\right) =
= 10−(10)321−10=∞\frac{\sqrt{\frac{1}{0}} - \left(\frac{1}{0}\right)^{\frac{3}{2}}}{1 - \sqrt{\frac{1}{0}}} = \infty

The final answer:
lim⁡x→∞(x+x)=∞\lim_{x \to \infty}\left(\sqrt{x} + x\right) = \infty
Lopital's rule
There is no sense to apply Lopital's rule to this function since there is no indeterminateness of 0/0 or oo/oo type
The graph
02468-8-6-4-2-1010020
Plot the graph
Rapid solution [src] 
oo
∞\infty
Expand and simplify
Other limits x→0, -oo, +oo, 1
lim⁡x→∞(x+x)=∞\lim_{x \to \infty}\left(\sqrt{x} + x\right) = \infty
lim⁡x→0−(x+x)=0\lim_{x \to 0^-}\left(\sqrt{x} + x\right) = 0
More at x→0 from the left
lim⁡x→0+(x+x)=0\lim_{x \to 0^+}\left(\sqrt{x} + x\right) = 0
More at x→0 from the right
lim⁡x→1−(x+x)=2\lim_{x \to 1^-}\left(\sqrt{x} + x\right) = 2
More at x→1 from the left
lim⁡x→1+(x+x)=2\lim_{x \to 1^+}\left(\sqrt{x} + x\right) = 2
More at x→1 from the right
lim⁡x→−∞(x+x)=−∞\lim_{x \to -\infty}\left(\sqrt{x} + x\right) = -\infty
More at x→-oo
The graph
Limit of the function x+sqrt(x)
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