Mister Exam
Other calculators:
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How to use it?
- Limit of the function:
-
Limit of sin(3)^2/x
-
Limit of x^4+2*cos(x)^2
-
Limit of x*2^x*3^(-x)
-
Limit of log(1+3*x^2)/(x^3-5*x^2)
- Integral of d{x}:
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1+x
- Expression:
- 1+x
- Derivative of:
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1+x
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Identical expressions
- one +x
- 1 plus x
- one plus x
-
Similar expressions
- 1-x
Limit of the function 1+x
The solution
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lim (1 + x) x->oo
$$\lim_{x \to \infty}\left(x + 1\right)$$
Limit(1 + x, x, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{x \to \infty}\left(x + 1\right)$$
Let's divide numerator and denominator by x:
$$\lim_{x \to \infty}\left(x + 1\right)$$ =
$$\lim_{x \to \infty}\left(\frac{1 + \frac{1}{x}}{\frac{1}{x}}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty}\left(\frac{1 + \frac{1}{x}}{\frac{1}{x}}\right) = \lim_{u \to 0^+}\left(\frac{u + 1}{u}\right)$$
=
$$\frac{1}{0} = \infty$$
The final answer:
$$\lim_{x \to \infty}\left(x + 1\right) = \infty$$
$$\lim_{x \to \infty}\left(x + 1\right)$$
Let's divide numerator and denominator by x:
$$\lim_{x \to \infty}\left(x + 1\right)$$ =
$$\lim_{x \to \infty}\left(\frac{1 + \frac{1}{x}}{\frac{1}{x}}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty}\left(\frac{1 + \frac{1}{x}}{\frac{1}{x}}\right) = \lim_{u \to 0^+}\left(\frac{u + 1}{u}\right)$$
=
$$\frac{1}{0} = \infty$$
The final answer:
$$\lim_{x \to \infty}\left(x + 1\right) = \infty$$
Lopital's rule
There is no sense to apply Lopital's rule to this function since there is no indeterminateness of 0/0 or oo/oo type
The graph
Other limits x→0, -oo, +oo, 1
$$\lim_{x \to \infty}\left(x + 1\right) = \infty$$
$$\lim_{x \to 0^-}\left(x + 1\right) = 1$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(x + 1\right) = 1$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(x + 1\right) = 2$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(x + 1\right) = 2$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(x + 1\right) = -\infty$$
More at x→-oo
$$\lim_{x \to 0^-}\left(x + 1\right) = 1$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(x + 1\right) = 1$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(x + 1\right) = 2$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(x + 1\right) = 2$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(x + 1\right) = -\infty$$
More at x→-oo
The graph