Mister Exam
Other calculators:
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How to use it?
- Limit of the function:
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Limit of (-5+2*x+3*x^4)/(7+x+2*x^2)
-
Limit of (5+x^2-6*x)/(-1+x^2)
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Limit of ((2+x)/(-2+x))^x
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Limit of (-3+sqrt(1+4*x))/(-8+x^3)
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Identical expressions
- ((two +x)/(- two +x))^x
- ((2 plus x) divide by ( minus 2 plus x)) to the power of x
- ((two plus x) divide by ( minus two plus x)) to the power of x
- ((2+x)/(-2+x))x
- 2+x/-2+xx
- 2+x/-2+x^x
- ((2+x) divide by (-2+x))^x
-
Similar expressions
- ((2+x)/(2+x))^x
- ((2-x)/(-2+x))^x
- ((2+x)/(-2-x))^x
Limit of the function ((2+x)/(-2+x))^x
The solution
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x
/2 + x \
lim |------|
x->oo\-2 + x/
$$\lim_{x \to \infty} \left(\frac{x + 2}{x - 2}\right)^{x}$$
Limit(((2 + x)/(-2 + x))^x, x, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{x \to \infty} \left(\frac{x + 2}{x - 2}\right)^{x}$$
transform
$$\lim_{x \to \infty} \left(\frac{x + 2}{x - 2}\right)^{x}$$
=
$$\lim_{x \to \infty} \left(\frac{\left(x - 2\right) + 4}{x - 2}\right)^{x}$$
=
$$\lim_{x \to \infty} \left(\frac{x - 2}{x - 2} + \frac{4}{x - 2}\right)^{x}$$
=
$$\lim_{x \to \infty} \left(1 + \frac{4}{x - 2}\right)^{x}$$
=
do replacement
$$u = \frac{x - 2}{4}$$
then
$$\lim_{x \to \infty} \left(1 + \frac{4}{x - 2}\right)^{x}$$ =
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{4 u + 2}$$
=
$$\lim_{u \to \infty}\left(\left(1 + \frac{1}{u}\right)^{2} \left(1 + \frac{1}{u}\right)^{4 u}\right)$$
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{2} \lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{4 u}$$
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{4 u}$$
=
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{4}$$
The limit
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}$$
is second remarkable limit, is equal to e ~ 2.718281828459045
then
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{4} = e^{4}$$
The final answer:
$$\lim_{x \to \infty} \left(\frac{x + 2}{x - 2}\right)^{x} = e^{4}$$
$$\lim_{x \to \infty} \left(\frac{x + 2}{x - 2}\right)^{x}$$
transform
$$\lim_{x \to \infty} \left(\frac{x + 2}{x - 2}\right)^{x}$$
=
$$\lim_{x \to \infty} \left(\frac{\left(x - 2\right) + 4}{x - 2}\right)^{x}$$
=
$$\lim_{x \to \infty} \left(\frac{x - 2}{x - 2} + \frac{4}{x - 2}\right)^{x}$$
=
$$\lim_{x \to \infty} \left(1 + \frac{4}{x - 2}\right)^{x}$$
=
do replacement
$$u = \frac{x - 2}{4}$$
then
$$\lim_{x \to \infty} \left(1 + \frac{4}{x - 2}\right)^{x}$$ =
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{4 u + 2}$$
=
$$\lim_{u \to \infty}\left(\left(1 + \frac{1}{u}\right)^{2} \left(1 + \frac{1}{u}\right)^{4 u}\right)$$
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{2} \lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{4 u}$$
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{4 u}$$
=
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{4}$$
The limit
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}$$
is second remarkable limit, is equal to e ~ 2.718281828459045
then
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{4} = e^{4}$$
The final answer:
$$\lim_{x \to \infty} \left(\frac{x + 2}{x - 2}\right)^{x} = e^{4}$$
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
$$\lim_{x \to \infty} \left(\frac{x + 2}{x - 2}\right)^{x} = e^{4}$$
$$\lim_{x \to 0^-} \left(\frac{x + 2}{x - 2}\right)^{x} = 1$$
More at x→0 from the left
$$\lim_{x \to 0^+} \left(\frac{x + 2}{x - 2}\right)^{x} = 1$$
More at x→0 from the right
$$\lim_{x \to 1^-} \left(\frac{x + 2}{x - 2}\right)^{x} = -3$$
More at x→1 from the left
$$\lim_{x \to 1^+} \left(\frac{x + 2}{x - 2}\right)^{x} = -3$$
More at x→1 from the right
$$\lim_{x \to -\infty} \left(\frac{x + 2}{x - 2}\right)^{x} = e^{4}$$
More at x→-oo
$$\lim_{x \to 0^-} \left(\frac{x + 2}{x - 2}\right)^{x} = 1$$
More at x→0 from the left
$$\lim_{x \to 0^+} \left(\frac{x + 2}{x - 2}\right)^{x} = 1$$
More at x→0 from the right
$$\lim_{x \to 1^-} \left(\frac{x + 2}{x - 2}\right)^{x} = -3$$
More at x→1 from the left
$$\lim_{x \to 1^+} \left(\frac{x + 2}{x - 2}\right)^{x} = -3$$
More at x→1 from the right
$$\lim_{x \to -\infty} \left(\frac{x + 2}{x - 2}\right)^{x} = e^{4}$$
More at x→-oo
The graph