Mister Exam
Other calculators:
-
How to use it?
- Limit of the function:
-
Limit of 1/(-1+e^(2*x))
-
Limit of (1-cos(x))/sin(2*x)
-
Limit of (1-cos(7*x))/x^2
-
Limit of (26-18*x^5+24*x^4)/(1-x^4-24*x^5-14*x^2+15*x+25*x^3)
- Integral of d{x}:
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x^4/(9-x^2)
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Identical expressions
- x^ four /(nine -x^ two)
- x to the power of 4 divide by (9 minus x squared )
- x to the power of four divide by (nine minus x to the power of two)
- x4/(9-x2)
- x4/9-x2
- x⁴/(9-x²)
- x to the power of 4/(9-x to the power of 2)
- x^4/9-x^2
- x^4 divide by (9-x^2)
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Similar expressions
- x^4/(9+x^2)
-
What you mean?
- x^4/(9 - x^2)
- x^4*2/(9 - x)
- x^4/((9 - x)^2)
- x^4/9 - x^2
- x*4/(9 - x^2)
- x*4*2/(9 - x)
- x^4/9 - x^2
You entered:
x^4/(9-x^2)
What you mean?
Limit of the function x^4/(9-x^2)
The solution
You have entered
[src]
/ 4 \
| x |
lim |------|
x->oo| 2|
\9 - x /
$$\lim_{x \to \infty}\left(\frac{x^{4}}{- x^{2} + 9}\right)$$
Limit(x^4/(9 - x^2), x, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{x \to \infty}\left(\frac{x^{4}}{- x^{2} + 9}\right)$$
Let's divide numerator and denominator by x^4:
$$\lim_{x \to \infty}\left(\frac{x^{4}}{- x^{2} + 9}\right)$$ =
$$\lim_{x \to \infty} \frac{1}{- \frac{1}{x^{2}} + \frac{9}{x^{4}}}$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty} \frac{1}{- \frac{1}{x^{2}} + \frac{9}{x^{4}}} = \lim_{u \to 0^+} \frac{1}{9 u^{4} - u^{2}}$$
=
$$\frac{1}{- 0^{2} + 9 \cdot 0^{4}} = -\infty$$
The final answer:
$$\lim_{x \to \infty}\left(\frac{x^{4}}{- x^{2} + 9}\right) = -\infty$$
$$\lim_{x \to \infty}\left(\frac{x^{4}}{- x^{2} + 9}\right)$$
Let's divide numerator and denominator by x^4:
$$\lim_{x \to \infty}\left(\frac{x^{4}}{- x^{2} + 9}\right)$$ =
$$\lim_{x \to \infty} \frac{1}{- \frac{1}{x^{2}} + \frac{9}{x^{4}}}$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty} \frac{1}{- \frac{1}{x^{2}} + \frac{9}{x^{4}}} = \lim_{u \to 0^+} \frac{1}{9 u^{4} - u^{2}}$$
=
$$\frac{1}{- 0^{2} + 9 \cdot 0^{4}} = -\infty$$
The final answer:
$$\lim_{x \to \infty}\left(\frac{x^{4}}{- x^{2} + 9}\right) = -\infty$$
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{x \to \infty} x^{4} = \infty$$
and limit for the denominator is
$$\lim_{x \to \infty}\left(- x^{2} + 9\right) = -\infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to \infty}\left(\frac{x^{4}}{- x^{2} + 9}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} x^{4}}{\frac{d}{d x} \left(- x^{2} + 9\right)}\right)$$
=
$$\lim_{x \to \infty}\left(- 2 x^{2}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} 4 x^{3}}{\frac{d}{d x} \left(- 2 x\right)}\right)$$
=
$$\lim_{x \to \infty}\left(- 6 x^{2}\right)$$
=
$$\lim_{x \to \infty}\left(- 6 x^{2}\right)$$
=
$$-\infty$$
It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 2 time(s)
oo/-oo,
i.e. limit for the numerator is
$$\lim_{x \to \infty} x^{4} = \infty$$
and limit for the denominator is
$$\lim_{x \to \infty}\left(- x^{2} + 9\right) = -\infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to \infty}\left(\frac{x^{4}}{- x^{2} + 9}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} x^{4}}{\frac{d}{d x} \left(- x^{2} + 9\right)}\right)$$
=
$$\lim_{x \to \infty}\left(- 2 x^{2}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} 4 x^{3}}{\frac{d}{d x} \left(- 2 x\right)}\right)$$
=
$$\lim_{x \to \infty}\left(- 6 x^{2}\right)$$
=
$$\lim_{x \to \infty}\left(- 6 x^{2}\right)$$
=
$$-\infty$$
It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 2 time(s)
The graph
Other limits x→0, -oo, +oo, 1
$$\lim_{x \to \infty}\left(\frac{x^{4}}{- x^{2} + 9}\right) = -\infty$$
$$\lim_{x \to 0^-}\left(\frac{x^{4}}{- x^{2} + 9}\right) = 0$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{x^{4}}{- x^{2} + 9}\right) = 0$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{x^{4}}{- x^{2} + 9}\right) = \frac{1}{8}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{x^{4}}{- x^{2} + 9}\right) = \frac{1}{8}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{x^{4}}{- x^{2} + 9}\right) = -\infty$$
More at x→-oo
$$\lim_{x \to 0^-}\left(\frac{x^{4}}{- x^{2} + 9}\right) = 0$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{x^{4}}{- x^{2} + 9}\right) = 0$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{x^{4}}{- x^{2} + 9}\right) = \frac{1}{8}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{x^{4}}{- x^{2} + 9}\right) = \frac{1}{8}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{x^{4}}{- x^{2} + 9}\right) = -\infty$$
More at x→-oo
The graph