Mister Exam
Other calculators:
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How to use it?
- Limit of the function:
-
Limit of (-6+x^2-x)/(6+x^2-5*x)
-
Limit of log(1+x^2)/(-e^(-x)+cos(3*x))
-
Limit of -2+|-2+x|/x
-
Limit of (-4-8*x+8*x^2)/(3+8*x)
- Integral of d{x}:
-
x^2/(1+x^3)
- Derivative of:
- x^2/(1+x^3)
-
Identical expressions
- x^ two /(one +x^ three)
- x squared divide by (1 plus x cubed )
- x to the power of two divide by (one plus x to the power of three)
- x2/(1+x3)
- x2/1+x3
- x²/(1+x³)
- x to the power of 2/(1+x to the power of 3)
- x^2/1+x^3
- x^2 divide by (1+x^3)
-
Similar expressions
- (-5+x)^2/(1+x)^3
- (-1+x)^2/(1+x)^3
- x^2/(1-x^3)
Limit of the function x^2/(1+x^3)
The solution
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/ 2 \
| x |
lim |------|
x->-oo| 3|
\1 + x /
$$\lim_{x \to -\infty}\left(\frac{x^{2}}{x^{3} + 1}\right)$$
Limit(x^2/(1 + x^3), x, -oo)
Detail solution
Let's take the limit
$$\lim_{x \to -\infty}\left(\frac{x^{2}}{x^{3} + 1}\right)$$
Let's divide numerator and denominator by x^3:
$$\lim_{x \to -\infty}\left(\frac{x^{2}}{x^{3} + 1}\right)$$ =
$$\lim_{x \to -\infty}\left(\frac{1}{x \left(1 + \frac{1}{x^{3}}\right)}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to -\infty}\left(\frac{1}{x \left(1 + \frac{1}{x^{3}}\right)}\right) = \lim_{u \to 0^+}\left(\frac{u}{u^{3} + 1}\right)$$
=
$$\frac{0}{0^{3} + 1} = 0$$
The final answer:
$$\lim_{x \to -\infty}\left(\frac{x^{2}}{x^{3} + 1}\right) = 0$$
$$\lim_{x \to -\infty}\left(\frac{x^{2}}{x^{3} + 1}\right)$$
Let's divide numerator and denominator by x^3:
$$\lim_{x \to -\infty}\left(\frac{x^{2}}{x^{3} + 1}\right)$$ =
$$\lim_{x \to -\infty}\left(\frac{1}{x \left(1 + \frac{1}{x^{3}}\right)}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to -\infty}\left(\frac{1}{x \left(1 + \frac{1}{x^{3}}\right)}\right) = \lim_{u \to 0^+}\left(\frac{u}{u^{3} + 1}\right)$$
=
$$\frac{0}{0^{3} + 1} = 0$$
The final answer:
$$\lim_{x \to -\infty}\left(\frac{x^{2}}{x^{3} + 1}\right) = 0$$
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{x \to -\infty} x^{2} = \infty$$
and limit for the denominator is
$$\lim_{x \to -\infty}\left(x^{3} + 1\right) = -\infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to -\infty}\left(\frac{x^{2}}{x^{3} + 1}\right)$$
=
$$\lim_{x \to -\infty}\left(\frac{\frac{d}{d x} x^{2}}{\frac{d}{d x} \left(x^{3} + 1\right)}\right)$$
=
$$\lim_{x \to -\infty}\left(\frac{2}{3 x}\right)$$
=
$$\lim_{x \to -\infty}\left(\frac{2}{3 x}\right)$$
=
$$0$$
It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 1 time(s)
oo/-oo,
i.e. limit for the numerator is
$$\lim_{x \to -\infty} x^{2} = \infty$$
and limit for the denominator is
$$\lim_{x \to -\infty}\left(x^{3} + 1\right) = -\infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to -\infty}\left(\frac{x^{2}}{x^{3} + 1}\right)$$
=
$$\lim_{x \to -\infty}\left(\frac{\frac{d}{d x} x^{2}}{\frac{d}{d x} \left(x^{3} + 1\right)}\right)$$
=
$$\lim_{x \to -\infty}\left(\frac{2}{3 x}\right)$$
=
$$\lim_{x \to -\infty}\left(\frac{2}{3 x}\right)$$
=
$$0$$
It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 1 time(s)
The graph
Other limits x→0, -oo, +oo, 1
$$\lim_{x \to -\infty}\left(\frac{x^{2}}{x^{3} + 1}\right) = 0$$
$$\lim_{x \to \infty}\left(\frac{x^{2}}{x^{3} + 1}\right) = 0$$
More at x→oo
$$\lim_{x \to 0^-}\left(\frac{x^{2}}{x^{3} + 1}\right) = 0$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{x^{2}}{x^{3} + 1}\right) = 0$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{x^{2}}{x^{3} + 1}\right) = \frac{1}{2}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{x^{2}}{x^{3} + 1}\right) = \frac{1}{2}$$
More at x→1 from the right
$$\lim_{x \to \infty}\left(\frac{x^{2}}{x^{3} + 1}\right) = 0$$
More at x→oo
$$\lim_{x \to 0^-}\left(\frac{x^{2}}{x^{3} + 1}\right) = 0$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{x^{2}}{x^{3} + 1}\right) = 0$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{x^{2}}{x^{3} + 1}\right) = \frac{1}{2}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{x^{2}}{x^{3} + 1}\right) = \frac{1}{2}$$
More at x→1 from the right
The graph