Mister Exam
Other calculators:
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How to use it?
- Limit of the function:
-
Limit of (3+2*x^2+5*x)/(7+3*x+5*x^2)
-
Limit of (1+x-3*x^3+2*x^2)/(-1+x^3)
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Limit of (2-x)^(1/(-1+x))
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Limit of ((-1+2*x)/(2+2*x))^x
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Identical expressions
- (two -x)^(one /(- one +x))
- (2 minus x) to the power of (1 divide by ( minus 1 plus x))
- (two minus x) to the power of (one divide by ( minus one plus x))
- (2-x)(1/(-1+x))
- 2-x1/-1+x
- 2-x^1/-1+x
- (2-x)^(1 divide by (-1+x))
-
Similar expressions
- (2-x)^(1/(-1-x))
- (2+x)^(1/(-1+x))
- (2-x)^(1/(1+x))
Limit of the function (2-x)^(1/(-1+x))
The solution
You have entered
[src]
1
------
-1 + x
lim (2 - x)
x->1+
$$\lim_{x \to 1^+} \left(2 - x\right)^{\frac{1}{x - 1}}$$
Limit((2 - x)^(1/(-1 + x)), x, 1)
Detail solution
Let's take the limit
$$\lim_{x \to 1^+} \left(2 - x\right)^{\frac{1}{x - 1}}$$
transform
do replacement
$$u = \frac{1}{1 - x}$$
then
$$\lim_{x \to 1^+} \left(1 + \frac{1}{\frac{1}{1 - x}}\right)^{\frac{1}{x - 1}}$$ =
=
$$\lim_{u \to 1^+} \left(2 - \frac{u - 1}{u}\right)^{\frac{1}{-1 + \frac{u - 1}{u}}}$$
=
$$\lim_{u \to 1^+} \left(1 + \frac{1}{u}\right)^{\frac{1}{-1 + \frac{u - 1}{u}}}$$
=
$$\left(\left(\lim_{u \to 1^+} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{\frac{1}{u \left(-1 + \frac{u - 1}{u}\right)}}$$
The limit
$$\lim_{u \to 1^+} \left(1 + \frac{1}{u}\right)^{u}$$
is second remarkable limit, is equal to e ~ 2.718281828459045
then
$$\left(\left(\lim_{u \to 1^+} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{\frac{1}{u \left(-1 + \frac{u - 1}{u}\right)}} = e^{\frac{1}{u \left(-1 + \frac{u - 1}{u}\right)}}$$
The final answer:
$$\lim_{x \to 1^+} \left(2 - x\right)^{\frac{1}{x - 1}} = e^{-1}$$
$$\lim_{x \to 1^+} \left(2 - x\right)^{\frac{1}{x - 1}}$$
transform
do replacement
$$u = \frac{1}{1 - x}$$
then
$$\lim_{x \to 1^+} \left(1 + \frac{1}{\frac{1}{1 - x}}\right)^{\frac{1}{x - 1}}$$ =
=
$$\lim_{u \to 1^+} \left(2 - \frac{u - 1}{u}\right)^{\frac{1}{-1 + \frac{u - 1}{u}}}$$
=
$$\lim_{u \to 1^+} \left(1 + \frac{1}{u}\right)^{\frac{1}{-1 + \frac{u - 1}{u}}}$$
=
$$\left(\left(\lim_{u \to 1^+} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{\frac{1}{u \left(-1 + \frac{u - 1}{u}\right)}}$$
The limit
$$\lim_{u \to 1^+} \left(1 + \frac{1}{u}\right)^{u}$$
is second remarkable limit, is equal to e ~ 2.718281828459045
then
$$\left(\left(\lim_{u \to 1^+} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{\frac{1}{u \left(-1 + \frac{u - 1}{u}\right)}} = e^{\frac{1}{u \left(-1 + \frac{u - 1}{u}\right)}}$$
The final answer:
$$\lim_{x \to 1^+} \left(2 - x\right)^{\frac{1}{x - 1}} = e^{-1}$$
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
One‐sided limits
[src]
1
------
-1 + x
lim (2 - x)
x->1+
$$\lim_{x \to 1^+} \left(2 - x\right)^{\frac{1}{x - 1}}$$
-1 e
$$e^{-1}$$
= 0.367879441171442
1
------
-1 + x
lim (2 - x)
x->1-
$$\lim_{x \to 1^-} \left(2 - x\right)^{\frac{1}{x - 1}}$$
-1 e
$$e^{-1}$$
= 0.367879441171442
= 0.367879441171442
Other limits x→0, -oo, +oo, 1
$$\lim_{x \to 1^-} \left(2 - x\right)^{\frac{1}{x - 1}} = e^{-1}$$
More at x→1 from the left
$$\lim_{x \to 1^+} \left(2 - x\right)^{\frac{1}{x - 1}} = e^{-1}$$
$$\lim_{x \to \infty} \left(2 - x\right)^{\frac{1}{x - 1}} = 1$$
More at x→oo
$$\lim_{x \to 0^-} \left(2 - x\right)^{\frac{1}{x - 1}} = \frac{1}{2}$$
More at x→0 from the left
$$\lim_{x \to 0^+} \left(2 - x\right)^{\frac{1}{x - 1}} = \frac{1}{2}$$
More at x→0 from the right
$$\lim_{x \to -\infty} \left(2 - x\right)^{\frac{1}{x - 1}} = 1$$
More at x→-oo
More at x→1 from the left
$$\lim_{x \to 1^+} \left(2 - x\right)^{\frac{1}{x - 1}} = e^{-1}$$
$$\lim_{x \to \infty} \left(2 - x\right)^{\frac{1}{x - 1}} = 1$$
More at x→oo
$$\lim_{x \to 0^-} \left(2 - x\right)^{\frac{1}{x - 1}} = \frac{1}{2}$$
More at x→0 from the left
$$\lim_{x \to 0^+} \left(2 - x\right)^{\frac{1}{x - 1}} = \frac{1}{2}$$
More at x→0 from the right
$$\lim_{x \to -\infty} \left(2 - x\right)^{\frac{1}{x - 1}} = 1$$
More at x→-oo
The graph