Mister Exam
Other calculators:
-
How to use it?
- Limit of the function:
-
Limit of (1+5*x+9*x^2)/(-2-7*x+2*x^3+4*x^2)
-
Limit of (9^x-2^(3*x))/(-7*x+atan(2*x))
-
Limit of (-1+7*x^2)/(1-2*x+4*x^2+5*x^3)
-
Limit of 7*x^2/(-1+e^(3*x))
- Integral of d{x}:
- x/(1-3*x^2)
-
Identical expressions
- x/(one - three *x^ two)
- x divide by (1 minus 3 multiply by x squared )
- x divide by (one minus three multiply by x to the power of two)
- x/(1-3*x2)
- x/1-3*x2
- x/(1-3*x²)
- x/(1-3*x to the power of 2)
- x/(1-3x^2)
- x/(1-3x2)
- x/1-3x2
- x/1-3x^2
- x divide by (1-3*x^2)
-
Similar expressions
- x/(1+3*x^2)
Limit of the function x/(1-3*x^2)
The solution
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/ x \
lim |--------|
x->oo| 2|
\1 - 3*x /
$$\lim_{x \to \infty}\left(\frac{x}{1 - 3 x^{2}}\right)$$
Limit(x/(1 - 3*x^2), x, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{x \to \infty}\left(\frac{x}{1 - 3 x^{2}}\right)$$
Let's divide numerator and denominator by x^2:
$$\lim_{x \to \infty}\left(\frac{x}{1 - 3 x^{2}}\right)$$ =
$$\lim_{x \to \infty}\left(\frac{1}{x \left(-3 + \frac{1}{x^{2}}\right)}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty}\left(\frac{1}{x \left(-3 + \frac{1}{x^{2}}\right)}\right) = \lim_{u \to 0^+}\left(\frac{u}{u^{2} - 3}\right)$$
=
$$\frac{0}{-3 + 0^{2}} = 0$$
The final answer:
$$\lim_{x \to \infty}\left(\frac{x}{1 - 3 x^{2}}\right) = 0$$
$$\lim_{x \to \infty}\left(\frac{x}{1 - 3 x^{2}}\right)$$
Let's divide numerator and denominator by x^2:
$$\lim_{x \to \infty}\left(\frac{x}{1 - 3 x^{2}}\right)$$ =
$$\lim_{x \to \infty}\left(\frac{1}{x \left(-3 + \frac{1}{x^{2}}\right)}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty}\left(\frac{1}{x \left(-3 + \frac{1}{x^{2}}\right)}\right) = \lim_{u \to 0^+}\left(\frac{u}{u^{2} - 3}\right)$$
=
$$\frac{0}{-3 + 0^{2}} = 0$$
The final answer:
$$\lim_{x \to \infty}\left(\frac{x}{1 - 3 x^{2}}\right) = 0$$
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{x \to \infty} x = \infty$$
and limit for the denominator is
$$\lim_{x \to \infty}\left(1 - 3 x^{2}\right) = -\infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to \infty}\left(\frac{x}{1 - 3 x^{2}}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} x}{\frac{d}{d x} \left(1 - 3 x^{2}\right)}\right)$$
=
$$\lim_{x \to \infty}\left(- \frac{1}{6 x}\right)$$
=
$$\lim_{x \to \infty}\left(- \frac{1}{6 x}\right)$$
=
$$0$$
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
$$\lim_{x \to \infty} x = \infty$$
and limit for the denominator is
$$\lim_{x \to \infty}\left(1 - 3 x^{2}\right) = -\infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to \infty}\left(\frac{x}{1 - 3 x^{2}}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} x}{\frac{d}{d x} \left(1 - 3 x^{2}\right)}\right)$$
=
$$\lim_{x \to \infty}\left(- \frac{1}{6 x}\right)$$
=
$$\lim_{x \to \infty}\left(- \frac{1}{6 x}\right)$$
=
$$0$$
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
$$\lim_{x \to \infty}\left(\frac{x}{1 - 3 x^{2}}\right) = 0$$
$$\lim_{x \to 0^-}\left(\frac{x}{1 - 3 x^{2}}\right) = 0$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{x}{1 - 3 x^{2}}\right) = 0$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{x}{1 - 3 x^{2}}\right) = - \frac{1}{2}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{x}{1 - 3 x^{2}}\right) = - \frac{1}{2}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{x}{1 - 3 x^{2}}\right) = 0$$
More at x→-oo
$$\lim_{x \to 0^-}\left(\frac{x}{1 - 3 x^{2}}\right) = 0$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{x}{1 - 3 x^{2}}\right) = 0$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{x}{1 - 3 x^{2}}\right) = - \frac{1}{2}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{x}{1 - 3 x^{2}}\right) = - \frac{1}{2}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{x}{1 - 3 x^{2}}\right) = 0$$
More at x→-oo
The graph