Mister Exam
Other calculators:
-
How to use it?
- Limit of the function:
-
Limit of ((-2+3*x)/(1+3*x))^(2*x)
-
Limit of ((-4+3*x)/(2+3*x))^(2*x)
-
Limit of (1-cos(4*x))/(1-cos(8*x))
-
Limit of (1+x^2-4*x)/(1+2*x)
-
Identical expressions
- (- eight +x^ two + two *x)/(eight -x^ three)
- ( minus 8 plus x squared plus 2 multiply by x) divide by (8 minus x cubed )
- ( minus eight plus x to the power of two plus two multiply by x) divide by (eight minus x to the power of three)
- (-8+x2+2*x)/(8-x3)
- -8+x2+2*x/8-x3
- (-8+x²+2*x)/(8-x³)
- (-8+x to the power of 2+2*x)/(8-x to the power of 3)
- (-8+x^2+2x)/(8-x^3)
- (-8+x2+2x)/(8-x3)
- -8+x2+2x/8-x3
- -8+x^2+2x/8-x^3
- (-8+x^2+2*x) divide by (8-x^3)
-
Similar expressions
- (-8+x^2-2*x)/(8-x^3)
- (-8+x^2+2*x)/(8+x^3)
- (8+x^2+2*x)/(8-x^3)
- (-8-x^2+2*x)/(8-x^3)
Limit of the function (-8+x^2+2*x)/(8-x^3)
The solution
You have entered
[src]
/ 2 \
|-8 + x + 2*x|
lim |-------------|
x->2+| 3 |
\ 8 - x /
$$\lim_{x \to 2^+}\left(\frac{2 x + \left(x^{2} - 8\right)}{8 - x^{3}}\right)$$
Limit((-8 + x^2 + 2*x)/(8 - x^3), x, 2)
Detail solution
Let's take the limit
$$\lim_{x \to 2^+}\left(\frac{2 x + \left(x^{2} - 8\right)}{8 - x^{3}}\right)$$
transform
$$\lim_{x \to 2^+}\left(\frac{2 x + \left(x^{2} - 8\right)}{8 - x^{3}}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{\left(x - 2\right) \left(x + 4\right)}{\left(-1\right) \left(x - 2\right) \left(x^{2} + 2 x + 4\right)}\right)$$
=
$$\lim_{x \to 2^+}\left(- \frac{x + 4}{x^{2} + 2 x + 4}\right) = $$
$$- \frac{2 + 4}{4 + 2^{2} + 2 \cdot 2} = $$
The final answer:
$$\lim_{x \to 2^+}\left(\frac{2 x + \left(x^{2} - 8\right)}{8 - x^{3}}\right) = - \frac{1}{2}$$
$$\lim_{x \to 2^+}\left(\frac{2 x + \left(x^{2} - 8\right)}{8 - x^{3}}\right)$$
transform
$$\lim_{x \to 2^+}\left(\frac{2 x + \left(x^{2} - 8\right)}{8 - x^{3}}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{\left(x - 2\right) \left(x + 4\right)}{\left(-1\right) \left(x - 2\right) \left(x^{2} + 2 x + 4\right)}\right)$$
=
$$\lim_{x \to 2^+}\left(- \frac{x + 4}{x^{2} + 2 x + 4}\right) = $$
$$- \frac{2 + 4}{4 + 2^{2} + 2 \cdot 2} = $$
= -1/2
The final answer:
$$\lim_{x \to 2^+}\left(\frac{2 x + \left(x^{2} - 8\right)}{8 - x^{3}}\right) = - \frac{1}{2}$$
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{x \to 2^+}\left(x^{2} + 2 x - 8\right) = 0$$
and limit for the denominator is
$$\lim_{x \to 2^+}\left(8 - x^{3}\right) = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 2^+}\left(\frac{2 x + \left(x^{2} - 8\right)}{8 - x^{3}}\right)$$
=
Let's transform the function under the limit a few
$$\lim_{x \to 2^+}\left(\frac{x^{2} + 2 x - 8}{8 - x^{3}}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{\frac{d}{d x} \left(x^{2} + 2 x - 8\right)}{\frac{d}{d x} \left(8 - x^{3}\right)}\right)$$
=
$$\lim_{x \to 2^+}\left(- \frac{2 x + 2}{3 x^{2}}\right)$$
=
$$\lim_{x \to 2^+}\left(- \frac{x}{6} - \frac{1}{6}\right)$$
=
$$\lim_{x \to 2^+}\left(- \frac{x}{6} - \frac{1}{6}\right)$$
=
$$- \frac{1}{2}$$
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)
0/0,
i.e. limit for the numerator is
$$\lim_{x \to 2^+}\left(x^{2} + 2 x - 8\right) = 0$$
and limit for the denominator is
$$\lim_{x \to 2^+}\left(8 - x^{3}\right) = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 2^+}\left(\frac{2 x + \left(x^{2} - 8\right)}{8 - x^{3}}\right)$$
=
Let's transform the function under the limit a few
$$\lim_{x \to 2^+}\left(\frac{x^{2} + 2 x - 8}{8 - x^{3}}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{\frac{d}{d x} \left(x^{2} + 2 x - 8\right)}{\frac{d}{d x} \left(8 - x^{3}\right)}\right)$$
=
$$\lim_{x \to 2^+}\left(- \frac{2 x + 2}{3 x^{2}}\right)$$
=
$$\lim_{x \to 2^+}\left(- \frac{x}{6} - \frac{1}{6}\right)$$
=
$$\lim_{x \to 2^+}\left(- \frac{x}{6} - \frac{1}{6}\right)$$
=
$$- \frac{1}{2}$$
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
$$\lim_{x \to 2^-}\left(\frac{2 x + \left(x^{2} - 8\right)}{8 - x^{3}}\right) = - \frac{1}{2}$$
More at x→2 from the left
$$\lim_{x \to 2^+}\left(\frac{2 x + \left(x^{2} - 8\right)}{8 - x^{3}}\right) = - \frac{1}{2}$$
$$\lim_{x \to \infty}\left(\frac{2 x + \left(x^{2} - 8\right)}{8 - x^{3}}\right) = 0$$
More at x→oo
$$\lim_{x \to 0^-}\left(\frac{2 x + \left(x^{2} - 8\right)}{8 - x^{3}}\right) = -1$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{2 x + \left(x^{2} - 8\right)}{8 - x^{3}}\right) = -1$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{2 x + \left(x^{2} - 8\right)}{8 - x^{3}}\right) = - \frac{5}{7}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{2 x + \left(x^{2} - 8\right)}{8 - x^{3}}\right) = - \frac{5}{7}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{2 x + \left(x^{2} - 8\right)}{8 - x^{3}}\right) = 0$$
More at x→-oo
More at x→2 from the left
$$\lim_{x \to 2^+}\left(\frac{2 x + \left(x^{2} - 8\right)}{8 - x^{3}}\right) = - \frac{1}{2}$$
$$\lim_{x \to \infty}\left(\frac{2 x + \left(x^{2} - 8\right)}{8 - x^{3}}\right) = 0$$
More at x→oo
$$\lim_{x \to 0^-}\left(\frac{2 x + \left(x^{2} - 8\right)}{8 - x^{3}}\right) = -1$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{2 x + \left(x^{2} - 8\right)}{8 - x^{3}}\right) = -1$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{2 x + \left(x^{2} - 8\right)}{8 - x^{3}}\right) = - \frac{5}{7}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{2 x + \left(x^{2} - 8\right)}{8 - x^{3}}\right) = - \frac{5}{7}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{2 x + \left(x^{2} - 8\right)}{8 - x^{3}}\right) = 0$$
More at x→-oo
One‐sided limits
[src]
/ 2 \
|-8 + x + 2*x|
lim |-------------|
x->2+| 3 |
\ 8 - x /
$$\lim_{x \to 2^+}\left(\frac{2 x + \left(x^{2} - 8\right)}{8 - x^{3}}\right)$$
-1/2
$$- \frac{1}{2}$$
= -0.5
/ 2 \
|-8 + x + 2*x|
lim |-------------|
x->2-| 3 |
\ 8 - x /
$$\lim_{x \to 2^-}\left(\frac{2 x + \left(x^{2} - 8\right)}{8 - x^{3}}\right)$$
-1/2
$$- \frac{1}{2}$$
= -0.5
= -0.5
The graph