Mister Exam
Other calculators:
-
How to use it?
- Limit of the function:
-
Limit of (-2*x^2+4*x^3+5*x)/(3*x^2+7*x)
-
Limit of (-cos(5*x)+cos(3*x))/x^2
-
Limit of (-1+cos(7*x))/(-1+cos(3*x))
-
Limit of (16+x^2+10*x)/(-6+x^2-x)
- Graphing y =:
- 2*x/(1-x^2)
- Integral of d{x}:
- 2*x/(1-x^2)
- Derivative of:
-
2*x/(1-x^2)
-
Identical expressions
- two *x/(one -x^ two)
- 2 multiply by x divide by (1 minus x squared )
- two multiply by x divide by (one minus x to the power of two)
- 2*x/(1-x2)
- 2*x/1-x2
- 2*x/(1-x²)
- 2*x/(1-x to the power of 2)
- 2x/(1-x^2)
- 2x/(1-x2)
- 2x/1-x2
- 2x/1-x^2
- 2*x divide by (1-x^2)
-
Similar expressions
- 2*x/(1+x^2)
Limit of the function 2*x/(1-x^2)
The solution
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[src]
/ 2*x \
lim |------|
x->oo| 2|
\1 - x /
$$\lim_{x \to \infty}\left(\frac{2 x}{1 - x^{2}}\right)$$
Limit((2*x)/(1 - x^2), x, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{x \to \infty}\left(\frac{2 x}{1 - x^{2}}\right)$$
Let's divide numerator and denominator by x^2:
$$\lim_{x \to \infty}\left(\frac{2 x}{1 - x^{2}}\right)$$ =
$$\lim_{x \to \infty}\left(\frac{2 \frac{1}{x}}{-1 + \frac{1}{x^{2}}}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty}\left(\frac{2 \frac{1}{x}}{-1 + \frac{1}{x^{2}}}\right) = \lim_{u \to 0^+}\left(\frac{2 u}{u^{2} - 1}\right)$$
=
$$\frac{0 \cdot 2}{-1 + 0^{2}} = 0$$
The final answer:
$$\lim_{x \to \infty}\left(\frac{2 x}{1 - x^{2}}\right) = 0$$
$$\lim_{x \to \infty}\left(\frac{2 x}{1 - x^{2}}\right)$$
Let's divide numerator and denominator by x^2:
$$\lim_{x \to \infty}\left(\frac{2 x}{1 - x^{2}}\right)$$ =
$$\lim_{x \to \infty}\left(\frac{2 \frac{1}{x}}{-1 + \frac{1}{x^{2}}}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty}\left(\frac{2 \frac{1}{x}}{-1 + \frac{1}{x^{2}}}\right) = \lim_{u \to 0^+}\left(\frac{2 u}{u^{2} - 1}\right)$$
=
$$\frac{0 \cdot 2}{-1 + 0^{2}} = 0$$
The final answer:
$$\lim_{x \to \infty}\left(\frac{2 x}{1 - x^{2}}\right) = 0$$
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{x \to \infty}\left(2 x\right) = \infty$$
and limit for the denominator is
$$\lim_{x \to \infty}\left(1 - x^{2}\right) = -\infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to \infty}\left(\frac{2 x}{1 - x^{2}}\right)$$
=
Let's transform the function under the limit a few
$$\lim_{x \to \infty}\left(\frac{2 x}{1 - x^{2}}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} 2 x}{\frac{d}{d x} \left(1 - x^{2}\right)}\right)$$
=
$$\lim_{x \to \infty}\left(- \frac{1}{x}\right)$$
=
$$\lim_{x \to \infty}\left(- \frac{1}{x}\right)$$
=
$$0$$
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
$$\lim_{x \to \infty}\left(2 x\right) = \infty$$
and limit for the denominator is
$$\lim_{x \to \infty}\left(1 - x^{2}\right) = -\infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to \infty}\left(\frac{2 x}{1 - x^{2}}\right)$$
=
Let's transform the function under the limit a few
$$\lim_{x \to \infty}\left(\frac{2 x}{1 - x^{2}}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} 2 x}{\frac{d}{d x} \left(1 - x^{2}\right)}\right)$$
=
$$\lim_{x \to \infty}\left(- \frac{1}{x}\right)$$
=
$$\lim_{x \to \infty}\left(- \frac{1}{x}\right)$$
=
$$0$$
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
$$\lim_{x \to \infty}\left(\frac{2 x}{1 - x^{2}}\right) = 0$$
$$\lim_{x \to 0^-}\left(\frac{2 x}{1 - x^{2}}\right) = 0$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{2 x}{1 - x^{2}}\right) = 0$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{2 x}{1 - x^{2}}\right) = \infty$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{2 x}{1 - x^{2}}\right) = -\infty$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{2 x}{1 - x^{2}}\right) = 0$$
More at x→-oo
$$\lim_{x \to 0^-}\left(\frac{2 x}{1 - x^{2}}\right) = 0$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{2 x}{1 - x^{2}}\right) = 0$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{2 x}{1 - x^{2}}\right) = \infty$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{2 x}{1 - x^{2}}\right) = -\infty$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{2 x}{1 - x^{2}}\right) = 0$$
More at x→-oo
The graph