Mister Exam
Other calculators:
-
How to use it?
- Limit of the function:
-
Limit of 10+x^2+3*x^3+8*x
-
Limit of (-16+2^x)/(-1+5*sqrt(x)*(5-x))
-
Limit of (-4+x^2)/(-3+sqrt(1-4*x))
-
Limit of ((1+x)/(-2+x))^(3+x)
-
Identical expressions
- ((one +x)^ three -(- one +x)^ three)/(one +x^ two)
- ((1 plus x) cubed minus ( minus 1 plus x) cubed ) divide by (1 plus x squared )
- ((one plus x) to the power of three minus ( minus one plus x) to the power of three) divide by (one plus x to the power of two)
- ((1+x)3-(-1+x)3)/(1+x2)
- 1+x3--1+x3/1+x2
- ((1+x)³-(-1+x)³)/(1+x²)
- ((1+x) to the power of 3-(-1+x) to the power of 3)/(1+x to the power of 2)
- 1+x^3--1+x^3/1+x^2
- ((1+x)^3-(-1+x)^3) divide by (1+x^2)
-
Similar expressions
- ((1+x)^3-(1+x)^3)/(1+x^2)
- ((1+x)^3-(-1+x)^3)/(1-x^2)
- ((1-x)^3-(-1+x)^3)/(1+x^2)
- ((1+x)^3+(-1+x)^3)/(1+x^2)
- ((1+x)^3-(-1-x)^3)/(1+x^2)
Limit of the function ((1+x)^3-(-1+x)^3)/(1+x^2)
The solution
You have entered
[src]
/ 3 3\
|(1 + x) - (-1 + x) |
lim |--------------------|
x->oo| 2 |
\ 1 + x /
$$\lim_{x \to \infty}\left(\frac{- \left(x - 1\right)^{3} + \left(x + 1\right)^{3}}{x^{2} + 1}\right)$$
Limit(((1 + x)^3 - (-1 + x)^3)/(1 + x^2), x, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{x \to \infty}\left(\frac{- \left(x - 1\right)^{3} + \left(x + 1\right)^{3}}{x^{2} + 1}\right)$$
Let's divide numerator and denominator by x^2:
$$\lim_{x \to \infty}\left(\frac{- \left(x - 1\right)^{3} + \left(x + 1\right)^{3}}{x^{2} + 1}\right)$$ =
$$\lim_{x \to \infty}\left(\frac{6 + \frac{2}{x^{2}}}{1 + \frac{1}{x^{2}}}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty}\left(\frac{6 + \frac{2}{x^{2}}}{1 + \frac{1}{x^{2}}}\right) = \lim_{u \to 0^+}\left(\frac{2 u^{2} + 6}{u^{2} + 1}\right)$$
=
$$\frac{2 \cdot 0^{2} + 6}{0^{2} + 1} = 6$$
The final answer:
$$\lim_{x \to \infty}\left(\frac{- \left(x - 1\right)^{3} + \left(x + 1\right)^{3}}{x^{2} + 1}\right) = 6$$
$$\lim_{x \to \infty}\left(\frac{- \left(x - 1\right)^{3} + \left(x + 1\right)^{3}}{x^{2} + 1}\right)$$
Let's divide numerator and denominator by x^2:
$$\lim_{x \to \infty}\left(\frac{- \left(x - 1\right)^{3} + \left(x + 1\right)^{3}}{x^{2} + 1}\right)$$ =
$$\lim_{x \to \infty}\left(\frac{6 + \frac{2}{x^{2}}}{1 + \frac{1}{x^{2}}}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty}\left(\frac{6 + \frac{2}{x^{2}}}{1 + \frac{1}{x^{2}}}\right) = \lim_{u \to 0^+}\left(\frac{2 u^{2} + 6}{u^{2} + 1}\right)$$
=
$$\frac{2 \cdot 0^{2} + 6}{0^{2} + 1} = 6$$
The final answer:
$$\lim_{x \to \infty}\left(\frac{- \left(x - 1\right)^{3} + \left(x + 1\right)^{3}}{x^{2} + 1}\right) = 6$$
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{x \to \infty}\left(6 x^{2} + 2\right) = \infty$$
and limit for the denominator is
$$\lim_{x \to \infty}\left(x^{2} + 1\right) = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to \infty}\left(\frac{- \left(x - 1\right)^{3} + \left(x + 1\right)^{3}}{x^{2} + 1}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \left(6 x^{2} + 2\right)}{\frac{d}{d x} \left(x^{2} + 1\right)}\right)$$
=
$$\lim_{x \to \infty} 6$$
=
$$\lim_{x \to \infty} 6$$
=
$$6$$
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
$$\lim_{x \to \infty}\left(6 x^{2} + 2\right) = \infty$$
and limit for the denominator is
$$\lim_{x \to \infty}\left(x^{2} + 1\right) = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to \infty}\left(\frac{- \left(x - 1\right)^{3} + \left(x + 1\right)^{3}}{x^{2} + 1}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \left(6 x^{2} + 2\right)}{\frac{d}{d x} \left(x^{2} + 1\right)}\right)$$
=
$$\lim_{x \to \infty} 6$$
=
$$\lim_{x \to \infty} 6$$
=
$$6$$
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
$$\lim_{x \to \infty}\left(\frac{- \left(x - 1\right)^{3} + \left(x + 1\right)^{3}}{x^{2} + 1}\right) = 6$$
$$\lim_{x \to 0^-}\left(\frac{- \left(x - 1\right)^{3} + \left(x + 1\right)^{3}}{x^{2} + 1}\right) = 2$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{- \left(x - 1\right)^{3} + \left(x + 1\right)^{3}}{x^{2} + 1}\right) = 2$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{- \left(x - 1\right)^{3} + \left(x + 1\right)^{3}}{x^{2} + 1}\right) = 4$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{- \left(x - 1\right)^{3} + \left(x + 1\right)^{3}}{x^{2} + 1}\right) = 4$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{- \left(x - 1\right)^{3} + \left(x + 1\right)^{3}}{x^{2} + 1}\right) = 6$$
More at x→-oo
$$\lim_{x \to 0^-}\left(\frac{- \left(x - 1\right)^{3} + \left(x + 1\right)^{3}}{x^{2} + 1}\right) = 2$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{- \left(x - 1\right)^{3} + \left(x + 1\right)^{3}}{x^{2} + 1}\right) = 2$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{- \left(x - 1\right)^{3} + \left(x + 1\right)^{3}}{x^{2} + 1}\right) = 4$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{- \left(x - 1\right)^{3} + \left(x + 1\right)^{3}}{x^{2} + 1}\right) = 4$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{- \left(x - 1\right)^{3} + \left(x + 1\right)^{3}}{x^{2} + 1}\right) = 6$$
More at x→-oo
The graph