Mister Exam
Other calculators:
-
How to use it?
- Limit of the function:
-
Limit of ((-1+2*x)/(2*x))^x
-
Limit of (36-32*x+7*x^4+12*x^6)/(36-32*x^5+7*x^6+12*x)
-
Limit of ((3+x)/(-4+x^2))^(6+3*x)
-
Limit of ((3+x)/(-2+x))^(-1+2*x)
- Graphing y =:
- sqrt(x)-x
- Derivative of:
- sqrt(x)-x
-
Identical expressions
- sqrt(x)-x
- square root of (x) minus x
- √(x)-x
- sqrtx-x
-
Similar expressions
- sqrt(x)*(x+sqrt(x+sqrt(x)))^(-x/2)
- sqrt(x)+x
Limit of the function sqrt(x)-x
The solution
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/ ___ \ lim \\/ x - x/ x->oo
$$\lim_{x \to \infty}\left(\sqrt{x} - x\right)$$
Limit(sqrt(x) - 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
$$\lim_{x \to \infty}\left(\sqrt{x} - x\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\left(\sqrt{x} - x\right) \left(\sqrt{x} + x\right)}{\sqrt{x} + x}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\left(\sqrt{x}\right)^{2} - x^{2}}{\sqrt{x} + x}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{- x^{2} + x}{\sqrt{x} + x}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{- x^{2} + x}{\sqrt{x} + x}\right)$$
Let's divide numerator and denominator by sqrt(x):
$$\lim_{x \to \infty}\left(\frac{- x^{\frac{3}{2}} + \sqrt{x}}{\sqrt{x} + 1}\right)$$ =
$$\lim_{x \to \infty}\left(\frac{- x^{\frac{3}{2}} + \sqrt{x}}{\sqrt{x} + 1}\right)$$ =
$$\lim_{x \to \infty}\left(\frac{- x^{\frac{3}{2}} + \sqrt{x}}{\sqrt{x} + 1}\right)$$
Do replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty}\left(\frac{- x^{\frac{3}{2}} + \sqrt{x}}{\sqrt{x} + 1}\right)$$ =
$$\lim_{u \to 0^+}\left(\frac{- \left(\frac{1}{u}\right)^{\frac{3}{2}} + \sqrt{\frac{1}{u}}}{\sqrt{\frac{1}{u}} + 1}\right)$$ =
= $$\frac{\sqrt{\frac{1}{0}} - \left(\frac{1}{0}\right)^{\frac{3}{2}}}{\sqrt{\frac{1}{0}} + 1} = -\infty$$
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
$$\lim_{x \to \infty}\left(\sqrt{x} - x\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\left(\sqrt{x} - x\right) \left(\sqrt{x} + x\right)}{\sqrt{x} + x}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\left(\sqrt{x}\right)^{2} - x^{2}}{\sqrt{x} + x}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{- x^{2} + x}{\sqrt{x} + x}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{- x^{2} + x}{\sqrt{x} + x}\right)$$
Let's divide numerator and denominator by sqrt(x):
$$\lim_{x \to \infty}\left(\frac{- x^{\frac{3}{2}} + \sqrt{x}}{\sqrt{x} + 1}\right)$$ =
$$\lim_{x \to \infty}\left(\frac{- x^{\frac{3}{2}} + \sqrt{x}}{\sqrt{x} + 1}\right)$$ =
$$\lim_{x \to \infty}\left(\frac{- x^{\frac{3}{2}} + \sqrt{x}}{\sqrt{x} + 1}\right)$$
Do replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty}\left(\frac{- x^{\frac{3}{2}} + \sqrt{x}}{\sqrt{x} + 1}\right)$$ =
$$\lim_{u \to 0^+}\left(\frac{- \left(\frac{1}{u}\right)^{\frac{3}{2}} + \sqrt{\frac{1}{u}}}{\sqrt{\frac{1}{u}} + 1}\right)$$ =
= $$\frac{\sqrt{\frac{1}{0}} - \left(\frac{1}{0}\right)^{\frac{3}{2}}}{\sqrt{\frac{1}{0}} + 1} = -\infty$$
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
$$\lim_{x \to 1^-}\left(\sqrt{x} - x\right) = 0$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\sqrt{x} - x\right) = 0$$
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
$$\lim_{x \to 1^-}\left(\sqrt{x} - x\right) = 0$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\sqrt{x} - x\right) = 0$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\sqrt{x} - x\right) = \infty$$
More at x→-oo
The graph