Mister Exam
Other calculators:
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How to use it?
- Limit of the function:
-
Limit of (3-3*x^2+4*x^4+6*x^3)/(2*x^2+7*x^4)
-
Limit of ((5+4*x)/(-1+5*x))^(1+3*x)
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Limit of (1+x^3-2*x)/(8+x^10-9*x)
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Limit of -2+x^2/3
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Identical expressions
- - two +x^ two / three
- minus 2 plus x squared divide by 3
- minus two plus x to the power of two divide by three
- -2+x2/3
- -2+x²/3
- -2+x to the power of 2/3
- -2+x^2 divide by 3
-
Similar expressions
- -2-x^2/3
- 2+x^2/3
- (1+x)^(2/3)-(-2+x)^(2/3)
Limit of the function -2+x^2/3
The solution
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/ 2\
| x |
lim |-2 + --|
x->oo\ 3 /
$$\lim_{x \to \infty}\left(\frac{x^{2}}{3} - 2\right)$$
Limit(-2 + x^2/3, x, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{x \to \infty}\left(\frac{x^{2}}{3} - 2\right)$$
Let's divide numerator and denominator by x^2:
$$\lim_{x \to \infty}\left(\frac{x^{2}}{3} - 2\right)$$ =
$$\lim_{x \to \infty}\left(\frac{\frac{1}{3} - \frac{2}{x^{2}}}{\frac{1}{x^{2}}}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty}\left(\frac{\frac{1}{3} - \frac{2}{x^{2}}}{\frac{1}{x^{2}}}\right) = \lim_{u \to 0^+}\left(\frac{\frac{1}{3} - 2 u^{2}}{u^{2}}\right)$$
=
$$\frac{\frac{1}{3} - 2 \cdot 0^{2}}{0} = \infty$$
The final answer:
$$\lim_{x \to \infty}\left(\frac{x^{2}}{3} - 2\right) = \infty$$
$$\lim_{x \to \infty}\left(\frac{x^{2}}{3} - 2\right)$$
Let's divide numerator and denominator by x^2:
$$\lim_{x \to \infty}\left(\frac{x^{2}}{3} - 2\right)$$ =
$$\lim_{x \to \infty}\left(\frac{\frac{1}{3} - \frac{2}{x^{2}}}{\frac{1}{x^{2}}}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty}\left(\frac{\frac{1}{3} - \frac{2}{x^{2}}}{\frac{1}{x^{2}}}\right) = \lim_{u \to 0^+}\left(\frac{\frac{1}{3} - 2 u^{2}}{u^{2}}\right)$$
=
$$\frac{\frac{1}{3} - 2 \cdot 0^{2}}{0} = \infty$$
The final answer:
$$\lim_{x \to \infty}\left(\frac{x^{2}}{3} - 2\right) = \infty$$
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
$$\lim_{x \to \infty}\left(\frac{x^{2}}{3} - 2\right) = \infty$$
$$\lim_{x \to 0^-}\left(\frac{x^{2}}{3} - 2\right) = -2$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{x^{2}}{3} - 2\right) = -2$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{x^{2}}{3} - 2\right) = - \frac{5}{3}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{x^{2}}{3} - 2\right) = - \frac{5}{3}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{x^{2}}{3} - 2\right) = \infty$$
More at x→-oo
$$\lim_{x \to 0^-}\left(\frac{x^{2}}{3} - 2\right) = -2$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{x^{2}}{3} - 2\right) = -2$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{x^{2}}{3} - 2\right) = - \frac{5}{3}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{x^{2}}{3} - 2\right) = - \frac{5}{3}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{x^{2}}{3} - 2\right) = \infty$$
More at x→-oo
The graph