Mister Exam
Other calculators:
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How to use it?
- Limit of the function:
-
Limit of (-15+x^2-2*x)/(-15-7*x+2*x^2)
-
Limit of (-1+e^(5*x))/sin(2*x)
-
Limit of (-4-8*x+8*x^2)/(3+8*x+9*x^2)
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Limit of (5-x)^(2/(-4+x))
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Identical expressions
- (five -x)^(two /(- four +x))
- (5 minus x) to the power of (2 divide by ( minus 4 plus x))
- (five minus x) to the power of (two divide by ( minus four plus x))
- (5-x)(2/(-4+x))
- 5-x2/-4+x
- 5-x^2/-4+x
- (5-x)^(2 divide by (-4+x))
-
Similar expressions
- (5-x)^(2/(-4-x))
- (5-x)^(2/(4+x))
- (5+x)^(2/(-4+x))
Limit of the function (5-x)^(2/(-4+x))
The solution
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[src]
2
------
-4 + x
lim (5 - x)
x->4+
$$\lim_{x \to 4^+} \left(5 - x\right)^{\frac{2}{x - 4}}$$
Limit((5 - x)^(2/(-4 + x)), x, 4)
Detail solution
Let's take the limit
$$\lim_{x \to 4^+} \left(5 - x\right)^{\frac{2}{x - 4}}$$
transform
do replacement
$$u = \frac{1}{4 - x}$$
then
$$\lim_{x \to 4^+} \left(1 + \frac{1}{\frac{1}{4 - x}}\right)^{\frac{2}{x - 4}}$$ =
=
$$\lim_{u \to 4^+} \left(1 + \frac{1}{u}\right)^{- 2 u}$$
=
$$\lim_{u \to 4^+} \left(1 + \frac{1}{u}\right)^{- 2 u}$$
=
$$\left(\left(\lim_{u \to 4^+} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{-2}$$
The limit
$$\lim_{u \to 4^+} \left(1 + \frac{1}{u}\right)^{u}$$
is second remarkable limit, is equal to e ~ 2.718281828459045
then
$$\left(\left(\lim_{u \to 4^+} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{-2} = e^{-2}$$
The final answer:
$$\lim_{x \to 4^+} \left(5 - x\right)^{\frac{2}{x - 4}} = e^{-2}$$
$$\lim_{x \to 4^+} \left(5 - x\right)^{\frac{2}{x - 4}}$$
transform
do replacement
$$u = \frac{1}{4 - x}$$
then
$$\lim_{x \to 4^+} \left(1 + \frac{1}{\frac{1}{4 - x}}\right)^{\frac{2}{x - 4}}$$ =
=
$$\lim_{u \to 4^+} \left(1 + \frac{1}{u}\right)^{- 2 u}$$
=
$$\lim_{u \to 4^+} \left(1 + \frac{1}{u}\right)^{- 2 u}$$
=
$$\left(\left(\lim_{u \to 4^+} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{-2}$$
The limit
$$\lim_{u \to 4^+} \left(1 + \frac{1}{u}\right)^{u}$$
is second remarkable limit, is equal to e ~ 2.718281828459045
then
$$\left(\left(\lim_{u \to 4^+} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{-2} = e^{-2}$$
The final answer:
$$\lim_{x \to 4^+} \left(5 - x\right)^{\frac{2}{x - 4}} = e^{-2}$$
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]
2
------
-4 + x
lim (5 - x)
x->4+
$$\lim_{x \to 4^+} \left(5 - x\right)^{\frac{2}{x - 4}}$$
-2 e
$$e^{-2}$$
= 0.135335283236613
2
------
-4 + x
lim (5 - x)
x->4-
$$\lim_{x \to 4^-} \left(5 - x\right)^{\frac{2}{x - 4}}$$
-2 e
$$e^{-2}$$
= 0.135335283236613
= 0.135335283236613
Other limits x→0, -oo, +oo, 1
$$\lim_{x \to 4^-} \left(5 - x\right)^{\frac{2}{x - 4}} = e^{-2}$$
More at x→4 from the left
$$\lim_{x \to 4^+} \left(5 - x\right)^{\frac{2}{x - 4}} = e^{-2}$$
$$\lim_{x \to \infty} \left(5 - x\right)^{\frac{2}{x - 4}} = 1$$
More at x→oo
$$\lim_{x \to 0^-} \left(5 - x\right)^{\frac{2}{x - 4}} = \frac{\sqrt{5}}{5}$$
More at x→0 from the left
$$\lim_{x \to 0^+} \left(5 - x\right)^{\frac{2}{x - 4}} = \frac{\sqrt{5}}{5}$$
More at x→0 from the right
$$\lim_{x \to 1^-} \left(5 - x\right)^{\frac{2}{x - 4}} = \frac{2^{\frac{2}{3}}}{4}$$
More at x→1 from the left
$$\lim_{x \to 1^+} \left(5 - x\right)^{\frac{2}{x - 4}} = \frac{2^{\frac{2}{3}}}{4}$$
More at x→1 from the right
$$\lim_{x \to -\infty} \left(5 - x\right)^{\frac{2}{x - 4}} = 1$$
More at x→-oo
More at x→4 from the left
$$\lim_{x \to 4^+} \left(5 - x\right)^{\frac{2}{x - 4}} = e^{-2}$$
$$\lim_{x \to \infty} \left(5 - x\right)^{\frac{2}{x - 4}} = 1$$
More at x→oo
$$\lim_{x \to 0^-} \left(5 - x\right)^{\frac{2}{x - 4}} = \frac{\sqrt{5}}{5}$$
More at x→0 from the left
$$\lim_{x \to 0^+} \left(5 - x\right)^{\frac{2}{x - 4}} = \frac{\sqrt{5}}{5}$$
More at x→0 from the right
$$\lim_{x \to 1^-} \left(5 - x\right)^{\frac{2}{x - 4}} = \frac{2^{\frac{2}{3}}}{4}$$
More at x→1 from the left
$$\lim_{x \to 1^+} \left(5 - x\right)^{\frac{2}{x - 4}} = \frac{2^{\frac{2}{3}}}{4}$$
More at x→1 from the right
$$\lim_{x \to -\infty} \left(5 - x\right)^{\frac{2}{x - 4}} = 1$$
More at x→-oo
The graph