Mister Exam
Other calculators:
-
How to use it?
- Limit of the function:
-
Limit of (-3+sqrt(1+4*x))/(-2+x)
-
Limit of 1/(-8+x)
-
Limit of (1-cos(6*x))/(1-cos(8*x))
-
Limit of (1-cos(3*x))/(-1+cos(7*x))
- Graphing y =:
- x/(2-x)
- Integral of d{x}:
-
x/(2-x)
-
Identical expressions
- x/(two -x)
- x divide by (2 minus x)
- x divide by (two minus x)
- x/2-x
- x divide by (2-x)
-
Similar expressions
- (1+x)/(2-x)
- (x^2-2*x)/(2-x)
- x*(x*e^x/2-x*e^2/2)
- x/(2+x)
Limit of the function x/(2-x)
The solution
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/ x \ lim |-----| x->-oo\2 - x/
$$\lim_{x \to -\infty}\left(\frac{x}{2 - x}\right)$$
Limit(x/(2 - x), x, -oo)
Detail solution
Let's take the limit
$$\lim_{x \to -\infty}\left(\frac{x}{2 - x}\right)$$
Let's divide numerator and denominator by x:
$$\lim_{x \to -\infty}\left(\frac{x}{2 - x}\right)$$ =
$$\lim_{x \to -\infty} \frac{1}{-1 + \frac{2}{x}}$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to -\infty} \frac{1}{-1 + \frac{2}{x}} = \lim_{u \to 0^+} \frac{1}{2 u - 1}$$
=
$$\frac{1}{-1 + 0 \cdot 2} = -1$$
The final answer:
$$\lim_{x \to -\infty}\left(\frac{x}{2 - x}\right) = -1$$
$$\lim_{x \to -\infty}\left(\frac{x}{2 - x}\right)$$
Let's divide numerator and denominator by x:
$$\lim_{x \to -\infty}\left(\frac{x}{2 - x}\right)$$ =
$$\lim_{x \to -\infty} \frac{1}{-1 + \frac{2}{x}}$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to -\infty} \frac{1}{-1 + \frac{2}{x}} = \lim_{u \to 0^+} \frac{1}{2 u - 1}$$
=
$$\frac{1}{-1 + 0 \cdot 2} = -1$$
The final answer:
$$\lim_{x \to -\infty}\left(\frac{x}{2 - x}\right) = -1$$
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{x \to -\infty} x = -\infty$$
and limit for the denominator is
$$\lim_{x \to -\infty}\left(2 - x\right) = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to -\infty}\left(\frac{x}{2 - x}\right)$$
=
$$\lim_{x \to -\infty}\left(\frac{\frac{d}{d x} x}{\frac{d}{d x} \left(2 - x\right)}\right)$$
=
$$\lim_{x \to -\infty} -1$$
=
$$\lim_{x \to -\infty} -1$$
=
$$-1$$
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
$$\lim_{x \to -\infty} x = -\infty$$
and limit for the denominator is
$$\lim_{x \to -\infty}\left(2 - x\right) = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to -\infty}\left(\frac{x}{2 - x}\right)$$
=
$$\lim_{x \to -\infty}\left(\frac{\frac{d}{d x} x}{\frac{d}{d x} \left(2 - x\right)}\right)$$
=
$$\lim_{x \to -\infty} -1$$
=
$$\lim_{x \to -\infty} -1$$
=
$$-1$$
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
$$\lim_{x \to -\infty}\left(\frac{x}{2 - x}\right) = -1$$
$$\lim_{x \to \infty}\left(\frac{x}{2 - x}\right) = -1$$
More at x→oo
$$\lim_{x \to 0^-}\left(\frac{x}{2 - x}\right) = 0$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{x}{2 - x}\right) = 0$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{x}{2 - x}\right) = 1$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{x}{2 - x}\right) = 1$$
More at x→1 from the right
$$\lim_{x \to \infty}\left(\frac{x}{2 - x}\right) = -1$$
More at x→oo
$$\lim_{x \to 0^-}\left(\frac{x}{2 - x}\right) = 0$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{x}{2 - x}\right) = 0$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{x}{2 - x}\right) = 1$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{x}{2 - x}\right) = 1$$
More at x→1 from the right
The graph