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