Mister Exam
Other calculators:
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How to use it?
- Limit of the function:
-
Limit of (-6+x^2-x)/(6+x^2-5*x)
-
Limit of log(1+x^2)/(-e^(-x)+cos(3*x))
-
Limit of -2+|-2+x|/x
-
Limit of (-4-8*x+8*x^2)/(3+8*x)
- Integral of d{x}:
-
(1+x)/x
- Graphing y =:
- (1+x)/x
-
Identical expressions
- (one +x)/x
- (1 plus x) divide by x
- (one plus x) divide by x
- 1+x/x
- (1+x) divide by x
-
Similar expressions
- (-1+x)/(x+x^2)
- (1-x)/x
Limit of the function (1+x)/x
The solution
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/1 + x\ lim |-----| x->oo\ x /
$$\lim_{x \to \infty}\left(\frac{x + 1}{x}\right)$$
Limit((1 + x)/x, x, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{x \to \infty}\left(\frac{x + 1}{x}\right)$$
Let's divide numerator and denominator by x:
$$\lim_{x \to \infty}\left(\frac{x + 1}{x}\right)$$ =
$$\lim_{x \to \infty}\left(\frac{1 + \frac{1}{x}}{1}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty}\left(\frac{1 + \frac{1}{x}}{1}\right) = \lim_{u \to 0^+}\left(u + 1\right)$$
=
$$1 = 1$$
The final answer:
$$\lim_{x \to \infty}\left(\frac{x + 1}{x}\right) = 1$$
$$\lim_{x \to \infty}\left(\frac{x + 1}{x}\right)$$
Let's divide numerator and denominator by x:
$$\lim_{x \to \infty}\left(\frac{x + 1}{x}\right)$$ =
$$\lim_{x \to \infty}\left(\frac{1 + \frac{1}{x}}{1}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty}\left(\frac{1 + \frac{1}{x}}{1}\right) = \lim_{u \to 0^+}\left(u + 1\right)$$
=
$$1 = 1$$
The final answer:
$$\lim_{x \to \infty}\left(\frac{x + 1}{x}\right) = 1$$
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
$$\lim_{x \to \infty} x = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to \infty}\left(\frac{x + 1}{x}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \left(x + 1\right)}{\frac{d}{d x} x}\right)$$
=
$$\lim_{x \to \infty} 1$$
=
$$\lim_{x \to \infty} 1$$
=
$$1$$
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
$$\lim_{x \to \infty} x = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to \infty}\left(\frac{x + 1}{x}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \left(x + 1\right)}{\frac{d}{d x} x}\right)$$
=
$$\lim_{x \to \infty} 1$$
=
$$\lim_{x \to \infty} 1$$
=
$$1$$
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
$$\lim_{x \to \infty}\left(\frac{x + 1}{x}\right) = 1$$
$$\lim_{x \to 0^-}\left(\frac{x + 1}{x}\right) = -\infty$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{x + 1}{x}\right) = \infty$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{x + 1}{x}\right) = 2$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{x + 1}{x}\right) = 2$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{x + 1}{x}\right) = 1$$
More at x→-oo
$$\lim_{x \to 0^-}\left(\frac{x + 1}{x}\right) = -\infty$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{x + 1}{x}\right) = \infty$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{x + 1}{x}\right) = 2$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{x + 1}{x}\right) = 2$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{x + 1}{x}\right) = 1$$
More at x→-oo
The graph