Mister Exam
Other calculators:
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How to use it?
- Limit of the function:
-
Limit of ((1+x)/(1+2*x))^x
-
Limit of (n/(1+n))^(5+3*n)
-
Limit of (9^x-8^x)/asin(3*x)
-
Limit of (-4+2*x+8*x^2)/(6+4*x)
- Derivative of:
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x+sqrt(x)
- Graphing y =:
- x+sqrt(x)
- Integral of d{x}:
-
x+sqrt(x)
-
Identical expressions
- x+sqrt(x)
- x plus square root of (x)
- x+√(x)
- x+sqrtx
-
Similar expressions
- x-sqrt(x)
Limit of the function x+sqrt(x)
The solution
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/ ___\ lim \x + \/ x / x->oo
$$\lim_{x \to \infty}\left(\sqrt{x} + x\right)$$
Limit(x + sqrt(x), x, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{x \to \infty}\left(\sqrt{x} + x\right)$$
Let's eliminate indeterminateness oo - oo
Multiply and divide by
$$\sqrt{x} - x$$
then
$$\lim_{x \to \infty}\left(\sqrt{x} + x\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\left(\sqrt{x} - x\right) \left(\sqrt{x} + x\right)}{\sqrt{x} - x}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\left(\sqrt{x}\right)^{2} - \left(- x\right)^{2}}{\sqrt{x} - x}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{- x^{2} + x}{\sqrt{x} - x}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{- x^{2} + x}{\sqrt{x} - x}\right)$$
Let's divide numerator and denominator by sqrt(x):
$$\lim_{x \to \infty}\left(\frac{- x^{\frac{3}{2}} + \sqrt{x}}{1 - \sqrt{x}}\right)$$ =
$$\lim_{x \to \infty}\left(\frac{- x^{\frac{3}{2}} + \sqrt{x}}{1 - \sqrt{x}}\right)$$ =
$$\lim_{x \to \infty}\left(\frac{- x^{\frac{3}{2}} + \sqrt{x}}{1 - \sqrt{x}}\right)$$
Do replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty}\left(\frac{- x^{\frac{3}{2}} + \sqrt{x}}{1 - \sqrt{x}}\right)$$ =
$$\lim_{u \to 0^+}\left(\frac{- \left(\frac{1}{u}\right)^{\frac{3}{2}} + \sqrt{\frac{1}{u}}}{1 - \sqrt{\frac{1}{u}}}\right)$$ =
= $$\frac{\sqrt{\frac{1}{0}} - \left(\frac{1}{0}\right)^{\frac{3}{2}}}{1 - \sqrt{\frac{1}{0}}} = \infty$$
The final answer:
$$\lim_{x \to \infty}\left(\sqrt{x} + x\right) = \infty$$
$$\lim_{x \to \infty}\left(\sqrt{x} + x\right)$$
Let's eliminate indeterminateness oo - oo
Multiply and divide by
$$\sqrt{x} - x$$
then
$$\lim_{x \to \infty}\left(\sqrt{x} + x\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\left(\sqrt{x} - x\right) \left(\sqrt{x} + x\right)}{\sqrt{x} - x}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\left(\sqrt{x}\right)^{2} - \left(- x\right)^{2}}{\sqrt{x} - x}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{- x^{2} + x}{\sqrt{x} - x}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{- x^{2} + x}{\sqrt{x} - x}\right)$$
Let's divide numerator and denominator by sqrt(x):
$$\lim_{x \to \infty}\left(\frac{- x^{\frac{3}{2}} + \sqrt{x}}{1 - \sqrt{x}}\right)$$ =
$$\lim_{x \to \infty}\left(\frac{- x^{\frac{3}{2}} + \sqrt{x}}{1 - \sqrt{x}}\right)$$ =
$$\lim_{x \to \infty}\left(\frac{- x^{\frac{3}{2}} + \sqrt{x}}{1 - \sqrt{x}}\right)$$
Do replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty}\left(\frac{- x^{\frac{3}{2}} + \sqrt{x}}{1 - \sqrt{x}}\right)$$ =
$$\lim_{u \to 0^+}\left(\frac{- \left(\frac{1}{u}\right)^{\frac{3}{2}} + \sqrt{\frac{1}{u}}}{1 - \sqrt{\frac{1}{u}}}\right)$$ =
= $$\frac{\sqrt{\frac{1}{0}} - \left(\frac{1}{0}\right)^{\frac{3}{2}}}{1 - \sqrt{\frac{1}{0}}} = \infty$$
The final answer:
$$\lim_{x \to \infty}\left(\sqrt{x} + x\right) = \infty$$
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
Other limits x→0, -oo, +oo, 1
$$\lim_{x \to \infty}\left(\sqrt{x} + x\right) = \infty$$
$$\lim_{x \to 0^-}\left(\sqrt{x} + x\right) = 0$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\sqrt{x} + x\right) = 0$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\sqrt{x} + x\right) = 2$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\sqrt{x} + x\right) = 2$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\sqrt{x} + x\right) = -\infty$$
More at x→-oo
$$\lim_{x \to 0^-}\left(\sqrt{x} + x\right) = 0$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\sqrt{x} + x\right) = 0$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\sqrt{x} + x\right) = 2$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\sqrt{x} + x\right) = 2$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\sqrt{x} + x\right) = -\infty$$
More at x→-oo
The graph