Mister Exam
Other calculators:
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How to use it?
- Limit of the function:
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Limit of ((2+x)/(-1+2*x))^x
-
Limit of ((2+x^2-2*x)/(2+x^2-4*x))^x
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Limit of (1+x^2-2*x)/(-5+x+4*x^2)
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Limit of (-25+x^2)/(-10+2*x)
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Identical expressions
- (- sixty-four +x^ three)/(- twenty-four +x^ two + two *x)
- ( minus 64 plus x cubed ) divide by ( minus 24 plus x squared plus 2 multiply by x)
- ( minus sixty minus four plus x to the power of three) divide by ( minus twenty minus four plus x to the power of two plus two multiply by x)
- (-64+x3)/(-24+x2+2*x)
- -64+x3/-24+x2+2*x
- (-64+x³)/(-24+x²+2*x)
- (-64+x to the power of 3)/(-24+x to the power of 2+2*x)
- (-64+x^3)/(-24+x^2+2x)
- (-64+x3)/(-24+x2+2x)
- -64+x3/-24+x2+2x
- -64+x^3/-24+x^2+2x
- (-64+x^3) divide by (-24+x^2+2*x)
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Similar expressions
- (-64+x^3)/(24+x^2+2*x)
- (-64-x^3)/(-24+x^2+2*x)
- (-64+x^3)/(-24+x^2-2*x)
- (64+x^3)/(-24+x^2+2*x)
- (-64+x^3)/(-24-x^2+2*x)
Limit of the function (-64+x^3)/(-24+x^2+2*x)
The solution
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[src]
/ 3 \
| -64 + x |
lim |--------------|
x->4+| 2 |
\-24 + x + 2*x/
$$\lim_{x \to 4^+}\left(\frac{x^{3} - 64}{2 x + \left(x^{2} - 24\right)}\right)$$
Limit((-64 + x^3)/(-24 + x^2 + 2*x), x, 4)
Detail solution
Let's take the limit
$$\lim_{x \to 4^+}\left(\frac{x^{3} - 64}{2 x + \left(x^{2} - 24\right)}\right)$$
transform
$$\lim_{x \to 4^+}\left(\frac{x^{3} - 64}{2 x + \left(x^{2} - 24\right)}\right)$$
=
$$\lim_{x \to 4^+}\left(\frac{\left(x - 4\right) \left(x^{2} + 4 x + 16\right)}{\left(x - 4\right) \left(x + 6\right)}\right)$$
=
$$\lim_{x \to 4^+}\left(\frac{x^{2} + 4 x + 16}{x + 6}\right) = $$
$$\frac{16 + 4^{2} + 4 \cdot 4}{4 + 6} = $$
The final answer:
$$\lim_{x \to 4^+}\left(\frac{x^{3} - 64}{2 x + \left(x^{2} - 24\right)}\right) = \frac{24}{5}$$
$$\lim_{x \to 4^+}\left(\frac{x^{3} - 64}{2 x + \left(x^{2} - 24\right)}\right)$$
transform
$$\lim_{x \to 4^+}\left(\frac{x^{3} - 64}{2 x + \left(x^{2} - 24\right)}\right)$$
=
$$\lim_{x \to 4^+}\left(\frac{\left(x - 4\right) \left(x^{2} + 4 x + 16\right)}{\left(x - 4\right) \left(x + 6\right)}\right)$$
=
$$\lim_{x \to 4^+}\left(\frac{x^{2} + 4 x + 16}{x + 6}\right) = $$
$$\frac{16 + 4^{2} + 4 \cdot 4}{4 + 6} = $$
= 24/5
The final answer:
$$\lim_{x \to 4^+}\left(\frac{x^{3} - 64}{2 x + \left(x^{2} - 24\right)}\right) = \frac{24}{5}$$
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{x \to 4^+}\left(x^{3} - 64\right) = 0$$
and limit for the denominator is
$$\lim_{x \to 4^+}\left(x^{2} + 2 x - 24\right) = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 4^+}\left(\frac{x^{3} - 64}{2 x + \left(x^{2} - 24\right)}\right)$$
=
Let's transform the function under the limit a few
$$\lim_{x \to 4^+}\left(\frac{x^{3} - 64}{x^{2} + 2 x - 24}\right)$$
=
$$\lim_{x \to 4^+}\left(\frac{\frac{d}{d x} \left(x^{3} - 64\right)}{\frac{d}{d x} \left(x^{2} + 2 x - 24\right)}\right)$$
=
$$\lim_{x \to 4^+}\left(\frac{3 x^{2}}{2 x + 2}\right)$$
=
$$\lim_{x \to 4^+}\left(\frac{48}{2 x + 2}\right)$$
=
$$\lim_{x \to 4^+}\left(\frac{48}{2 x + 2}\right)$$
=
$$\frac{24}{5}$$
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
$$\lim_{x \to 4^+}\left(x^{3} - 64\right) = 0$$
and limit for the denominator is
$$\lim_{x \to 4^+}\left(x^{2} + 2 x - 24\right) = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 4^+}\left(\frac{x^{3} - 64}{2 x + \left(x^{2} - 24\right)}\right)$$
=
Let's transform the function under the limit a few
$$\lim_{x \to 4^+}\left(\frac{x^{3} - 64}{x^{2} + 2 x - 24}\right)$$
=
$$\lim_{x \to 4^+}\left(\frac{\frac{d}{d x} \left(x^{3} - 64\right)}{\frac{d}{d x} \left(x^{2} + 2 x - 24\right)}\right)$$
=
$$\lim_{x \to 4^+}\left(\frac{3 x^{2}}{2 x + 2}\right)$$
=
$$\lim_{x \to 4^+}\left(\frac{48}{2 x + 2}\right)$$
=
$$\lim_{x \to 4^+}\left(\frac{48}{2 x + 2}\right)$$
=
$$\frac{24}{5}$$
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
One‐sided limits
[src]
/ 3 \
| -64 + x |
lim |--------------|
x->4+| 2 |
\-24 + x + 2*x/
$$\lim_{x \to 4^+}\left(\frac{x^{3} - 64}{2 x + \left(x^{2} - 24\right)}\right)$$
24/5
$$\frac{24}{5}$$
= 4.8
/ 3 \
| -64 + x |
lim |--------------|
x->4-| 2 |
\-24 + x + 2*x/
$$\lim_{x \to 4^-}\left(\frac{x^{3} - 64}{2 x + \left(x^{2} - 24\right)}\right)$$
24/5
$$\frac{24}{5}$$
= 4.8
= 4.8
Other limits x→0, -oo, +oo, 1
$$\lim_{x \to 4^-}\left(\frac{x^{3} - 64}{2 x + \left(x^{2} - 24\right)}\right) = \frac{24}{5}$$
More at x→4 from the left
$$\lim_{x \to 4^+}\left(\frac{x^{3} - 64}{2 x + \left(x^{2} - 24\right)}\right) = \frac{24}{5}$$
$$\lim_{x \to \infty}\left(\frac{x^{3} - 64}{2 x + \left(x^{2} - 24\right)}\right) = \infty$$
More at x→oo
$$\lim_{x \to 0^-}\left(\frac{x^{3} - 64}{2 x + \left(x^{2} - 24\right)}\right) = \frac{8}{3}$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{x^{3} - 64}{2 x + \left(x^{2} - 24\right)}\right) = \frac{8}{3}$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{x^{3} - 64}{2 x + \left(x^{2} - 24\right)}\right) = 3$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{x^{3} - 64}{2 x + \left(x^{2} - 24\right)}\right) = 3$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{x^{3} - 64}{2 x + \left(x^{2} - 24\right)}\right) = -\infty$$
More at x→-oo
More at x→4 from the left
$$\lim_{x \to 4^+}\left(\frac{x^{3} - 64}{2 x + \left(x^{2} - 24\right)}\right) = \frac{24}{5}$$
$$\lim_{x \to \infty}\left(\frac{x^{3} - 64}{2 x + \left(x^{2} - 24\right)}\right) = \infty$$
More at x→oo
$$\lim_{x \to 0^-}\left(\frac{x^{3} - 64}{2 x + \left(x^{2} - 24\right)}\right) = \frac{8}{3}$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{x^{3} - 64}{2 x + \left(x^{2} - 24\right)}\right) = \frac{8}{3}$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{x^{3} - 64}{2 x + \left(x^{2} - 24\right)}\right) = 3$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{x^{3} - 64}{2 x + \left(x^{2} - 24\right)}\right) = 3$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{x^{3} - 64}{2 x + \left(x^{2} - 24\right)}\right) = -\infty$$
More at x→-oo
The graph