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}:
- (1+x)/x^2
-
Identical expressions
- (one +x)/x^ two
- (1 plus x) divide by x squared
- (one plus x) divide by x to the power of two
- (1+x)/x2
- 1+x/x2
- (1+x)/x²
- (1+x)/x to the power of 2
- 1+x/x^2
- (1+x) divide by x^2
-
Similar expressions
- (-1+x)/x^2
- cos(x)^2*(-1+x)/x^2
- (1-x)/x^2
- (-x*e^x+log(1+x))/x^2
- -1+sqrt(2-x)*(-1+x)/x^2
- (-x+log(1+x))/x^2
Limit of the function (1+x)/x^2
The solution
You have entered
[src]
/1 + x\
lim |-----|
x->oo| 2 |
\ x /
$$\lim_{x \to \infty}\left(\frac{x + 1}{x^{2}}\right)$$
Limit((1 + x)/x^2, x, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{x \to \infty}\left(\frac{x + 1}{x^{2}}\right)$$
Let's divide numerator and denominator by x^2:
$$\lim_{x \to \infty}\left(\frac{x + 1}{x^{2}}\right)$$ =
$$\lim_{x \to \infty}\left(\frac{\frac{1}{x} + \frac{1}{x^{2}}}{1}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty}\left(\frac{\frac{1}{x} + \frac{1}{x^{2}}}{1}\right) = \lim_{u \to 0^+}\left(u^{2} + u\right)$$
=
$$0^{2} = 0$$
The final answer:
$$\lim_{x \to \infty}\left(\frac{x + 1}{x^{2}}\right) = 0$$
$$\lim_{x \to \infty}\left(\frac{x + 1}{x^{2}}\right)$$
Let's divide numerator and denominator by x^2:
$$\lim_{x \to \infty}\left(\frac{x + 1}{x^{2}}\right)$$ =
$$\lim_{x \to \infty}\left(\frac{\frac{1}{x} + \frac{1}{x^{2}}}{1}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty}\left(\frac{\frac{1}{x} + \frac{1}{x^{2}}}{1}\right) = \lim_{u \to 0^+}\left(u^{2} + u\right)$$
=
$$0^{2} = 0$$
The final answer:
$$\lim_{x \to \infty}\left(\frac{x + 1}{x^{2}}\right) = 0$$
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{x \to \infty}\left(x + 1\right) = \infty$$
and limit for the denominator is
$$\lim_{x \to \infty} x^{2} = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to \infty}\left(\frac{x + 1}{x^{2}}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \left(x + 1\right)}{\frac{d}{d x} x^{2}}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{1}{2 x}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{1}{2 x}\right)$$
=
$$0$$
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
$$\lim_{x \to \infty}\left(x + 1\right) = \infty$$
and limit for the denominator is
$$\lim_{x \to \infty} x^{2} = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to \infty}\left(\frac{x + 1}{x^{2}}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \left(x + 1\right)}{\frac{d}{d x} x^{2}}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{1}{2 x}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{1}{2 x}\right)$$
=
$$0$$
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
$$\lim_{x \to \infty}\left(\frac{x + 1}{x^{2}}\right) = 0$$
$$\lim_{x \to 0^-}\left(\frac{x + 1}{x^{2}}\right) = \infty$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{x + 1}{x^{2}}\right) = \infty$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{x + 1}{x^{2}}\right) = 2$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{x + 1}{x^{2}}\right) = 2$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{x + 1}{x^{2}}\right) = 0$$
More at x→-oo
$$\lim_{x \to 0^-}\left(\frac{x + 1}{x^{2}}\right) = \infty$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{x + 1}{x^{2}}\right) = \infty$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{x + 1}{x^{2}}\right) = 2$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{x + 1}{x^{2}}\right) = 2$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{x + 1}{x^{2}}\right) = 0$$
More at x→-oo
The graph