Mister Exam
Other calculators:
-
How to use it?
- Limit of the function:
-
Limit of (x+sqrt(10+x-x^2))/(sqrt(2-7*x)+2*x)
- Limit of (-1+x^m)/(-1+x)
- Limit of x*a^(-x)
-
Limit of -4+x+x^3
-
Identical expressions
- - four +x+x^ three
- minus 4 plus x plus x cubed
- minus four plus x plus x to the power of three
- -4+x+x3
- -4+x+x³
- -4+x+x to the power of 3
-
Similar expressions
- -4+x-x^3
- -4-x+x^3
- 4+x+x^3
Limit of the function -4+x+x^3
The solution
You have entered
[src]
/ 3\ lim \-4 + x + x / x->oo
$$\lim_{x \to \infty}\left(x^{3} + \left(x - 4\right)\right)$$
Limit(-4 + x + x^3, x, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{x \to \infty}\left(x^{3} + \left(x - 4\right)\right)$$
Let's divide numerator and denominator by x^3:
$$\lim_{x \to \infty}\left(x^{3} + \left(x - 4\right)\right)$$ =
$$\lim_{x \to \infty}\left(\frac{1 + \frac{1}{x^{2}} - \frac{4}{x^{3}}}{\frac{1}{x^{3}}}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty}\left(\frac{1 + \frac{1}{x^{2}} - \frac{4}{x^{3}}}{\frac{1}{x^{3}}}\right) = \lim_{u \to 0^+}\left(\frac{- 4 u^{3} + u^{2} + 1}{u^{3}}\right)$$
=
$$\frac{0^{2} - 4 \cdot 0^{3} + 1}{0} = \infty$$
The final answer:
$$\lim_{x \to \infty}\left(x^{3} + \left(x - 4\right)\right) = \infty$$
$$\lim_{x \to \infty}\left(x^{3} + \left(x - 4\right)\right)$$
Let's divide numerator and denominator by x^3:
$$\lim_{x \to \infty}\left(x^{3} + \left(x - 4\right)\right)$$ =
$$\lim_{x \to \infty}\left(\frac{1 + \frac{1}{x^{2}} - \frac{4}{x^{3}}}{\frac{1}{x^{3}}}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty}\left(\frac{1 + \frac{1}{x^{2}} - \frac{4}{x^{3}}}{\frac{1}{x^{3}}}\right) = \lim_{u \to 0^+}\left(\frac{- 4 u^{3} + u^{2} + 1}{u^{3}}\right)$$
=
$$\frac{0^{2} - 4 \cdot 0^{3} + 1}{0} = \infty$$
The final answer:
$$\lim_{x \to \infty}\left(x^{3} + \left(x - 4\right)\right) = \infty$$
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
$$\lim_{x \to \infty}\left(x^{3} + \left(x - 4\right)\right) = \infty$$
$$\lim_{x \to 0^-}\left(x^{3} + \left(x - 4\right)\right) = -4$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(x^{3} + \left(x - 4\right)\right) = -4$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(x^{3} + \left(x - 4\right)\right) = -2$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(x^{3} + \left(x - 4\right)\right) = -2$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(x^{3} + \left(x - 4\right)\right) = -\infty$$
More at x→-oo
$$\lim_{x \to 0^-}\left(x^{3} + \left(x - 4\right)\right) = -4$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(x^{3} + \left(x - 4\right)\right) = -4$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(x^{3} + \left(x - 4\right)\right) = -2$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(x^{3} + \left(x - 4\right)\right) = -2$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(x^{3} + \left(x - 4\right)\right) = -\infty$$
More at x→-oo
The graph