Mister Exam
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How to use it?
- Limit of the function:
- Limit of n2*(5/2+n/2)
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Limit of (-8+x^2+2*x)/(8-x^3)
-
Limit of (-2+x)/(-8+x^3)
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Limit of (3+x^2+2*x)/(4+2*x^2+3*x)
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Identical expressions
- (- two +x)/(- eight +x^ three)
- ( minus 2 plus x) divide by ( minus 8 plus x cubed )
- ( minus two plus x) divide by ( minus eight plus x to the power of three)
- (-2+x)/(-8+x3)
- -2+x/-8+x3
- (-2+x)/(-8+x³)
- (-2+x)/(-8+x to the power of 3)
- -2+x/-8+x^3
- (-2+x) divide by (-8+x^3)
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Similar expressions
- (-2-x)/(-8+x^3)
- (-2+x)/(-8-x^3)
- (2+x)/(-8+x^3)
- (-2+x)/(8+x^3)
Limit of the function (-2+x)/(-8+x^3)
The solution
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[src]
/ -2 + x\
lim |-------|
x->2+| 3|
\-8 + x /
$$\lim_{x \to 2^+}\left(\frac{x - 2}{x^{3} - 8}\right)$$
Limit((-2 + x)/(-8 + x^3), x, 2)
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{x \to 2^+}\left(x - 2\right) = 0$$
and limit for the denominator is
$$\lim_{x \to 2^+}\left(x^{3} - 8\right) = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 2^+}\left(\frac{x - 2}{x^{3} - 8}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{\frac{d}{d x} \left(x - 2\right)}{\frac{d}{d x} \left(x^{3} - 8\right)}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{1}{3 x^{2}}\right)$$
=
$$\lim_{x \to 2^+} \frac{1}{12}$$
=
$$\lim_{x \to 2^+} \frac{1}{12}$$
=
$$\frac{1}{12}$$
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
$$\lim_{x \to 2^+}\left(x - 2\right) = 0$$
and limit for the denominator is
$$\lim_{x \to 2^+}\left(x^{3} - 8\right) = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 2^+}\left(\frac{x - 2}{x^{3} - 8}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{\frac{d}{d x} \left(x - 2\right)}{\frac{d}{d x} \left(x^{3} - 8\right)}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{1}{3 x^{2}}\right)$$
=
$$\lim_{x \to 2^+} \frac{1}{12}$$
=
$$\lim_{x \to 2^+} \frac{1}{12}$$
=
$$\frac{1}{12}$$
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
$$\lim_{x \to 2^-}\left(\frac{x - 2}{x^{3} - 8}\right) = \frac{1}{12}$$
More at x→2 from the left
$$\lim_{x \to 2^+}\left(\frac{x - 2}{x^{3} - 8}\right) = \frac{1}{12}$$
$$\lim_{x \to \infty}\left(\frac{x - 2}{x^{3} - 8}\right) = 0$$
More at x→oo
$$\lim_{x \to 0^-}\left(\frac{x - 2}{x^{3} - 8}\right) = \frac{1}{4}$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{x - 2}{x^{3} - 8}\right) = \frac{1}{4}$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{x - 2}{x^{3} - 8}\right) = \frac{1}{7}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{x - 2}{x^{3} - 8}\right) = \frac{1}{7}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{x - 2}{x^{3} - 8}\right) = 0$$
More at x→-oo
More at x→2 from the left
$$\lim_{x \to 2^+}\left(\frac{x - 2}{x^{3} - 8}\right) = \frac{1}{12}$$
$$\lim_{x \to \infty}\left(\frac{x - 2}{x^{3} - 8}\right) = 0$$
More at x→oo
$$\lim_{x \to 0^-}\left(\frac{x - 2}{x^{3} - 8}\right) = \frac{1}{4}$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{x - 2}{x^{3} - 8}\right) = \frac{1}{4}$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{x - 2}{x^{3} - 8}\right) = \frac{1}{7}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{x - 2}{x^{3} - 8}\right) = \frac{1}{7}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{x - 2}{x^{3} - 8}\right) = 0$$
More at x→-oo
One‐sided limits
[src]
/ -2 + x\
lim |-------|
x->2+| 3|
\-8 + x /
$$\lim_{x \to 2^+}\left(\frac{x - 2}{x^{3} - 8}\right)$$
1/12
$$\frac{1}{12}$$
= 0.0833333333333333
/ -2 + x\
lim |-------|
x->2-| 3|
\-8 + x /
$$\lim_{x \to 2^-}\left(\frac{x - 2}{x^{3} - 8}\right)$$
1/12
$$\frac{1}{12}$$
= 0.0833333333333333
= 0.0833333333333333
The graph