Mister Exam
Other calculators:
-
How to use it?
- Limit of the function:
-
Limit of -2+x^3+6*x
-
Limit of ((-2+x)/x)^(2*x)
-
Limit of (-2+x)*log(-2+x)
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Limit of -6+x^2+5*x
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Identical expressions
- ((- two +x)/x)^(two *x)
- (( minus 2 plus x) divide by x) to the power of (2 multiply by x)
- (( minus two plus x) divide by x) to the power of (two multiply by x)
- ((-2+x)/x)(2*x)
- -2+x/x2*x
- ((-2+x)/x)^(2x)
- ((-2+x)/x)(2x)
- -2+x/x2x
- -2+x/x^2x
- ((-2+x) divide by x)^(2*x)
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Similar expressions
- ((-2-x)/x)^(2*x)
- ((2+x)/x)^(2*x)
Limit of the function ((-2+x)/x)^(2*x)
The solution
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2*x
/-2 + x\
lim |------|
x->oo\ x /
$$\lim_{x \to \infty} \left(\frac{x - 2}{x}\right)^{2 x}$$
Limit(((-2 + x)/x)^(2*x), x, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{x \to \infty} \left(\frac{x - 2}{x}\right)^{2 x}$$
transform
$$\lim_{x \to \infty} \left(\frac{x - 2}{x}\right)^{2 x}$$
=
$$\lim_{x \to \infty} \left(\frac{x - 2}{x}\right)^{2 x}$$
=
$$\lim_{x \to \infty} \left(- \frac{2}{x} + \frac{x}{x}\right)^{2 x}$$
=
$$\lim_{x \to \infty} \left(1 - \frac{2}{x}\right)^{2 x}$$
=
do replacement
$$u = \frac{x}{-2}$$
then
$$\lim_{x \to \infty} \left(1 - \frac{2}{x}\right)^{2 x}$$ =
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- 4 u}$$
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- 4 u}$$
=
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{-4}$$
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:
$$\lim_{x \to \infty} \left(\frac{x - 2}{x}\right)^{2 x} = e^{-4}$$
$$\lim_{x \to \infty} \left(\frac{x - 2}{x}\right)^{2 x}$$
transform
$$\lim_{x \to \infty} \left(\frac{x - 2}{x}\right)^{2 x}$$
=
$$\lim_{x \to \infty} \left(\frac{x - 2}{x}\right)^{2 x}$$
=
$$\lim_{x \to \infty} \left(- \frac{2}{x} + \frac{x}{x}\right)^{2 x}$$
=
$$\lim_{x \to \infty} \left(1 - \frac{2}{x}\right)^{2 x}$$
=
do replacement
$$u = \frac{x}{-2}$$
then
$$\lim_{x \to \infty} \left(1 - \frac{2}{x}\right)^{2 x}$$ =
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- 4 u}$$
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- 4 u}$$
=
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{-4}$$
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:
$$\lim_{x \to \infty} \left(\frac{x - 2}{x}\right)^{2 x} = e^{-4}$$
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
$$\lim_{x \to \infty} \left(\frac{x - 2}{x}\right)^{2 x} = e^{-4}$$
$$\lim_{x \to 0^-} \left(\frac{x - 2}{x}\right)^{2 x} = 1$$
More at x→0 from the left
$$\lim_{x \to 0^+} \left(\frac{x - 2}{x}\right)^{2 x} = 1$$
More at x→0 from the right
$$\lim_{x \to 1^-} \left(\frac{x - 2}{x}\right)^{2 x} = 1$$
More at x→1 from the left
$$\lim_{x \to 1^+} \left(\frac{x - 2}{x}\right)^{2 x} = 1$$
More at x→1 from the right
$$\lim_{x \to -\infty} \left(\frac{x - 2}{x}\right)^{2 x} = e^{-4}$$
More at x→-oo
$$\lim_{x \to 0^-} \left(\frac{x - 2}{x}\right)^{2 x} = 1$$
More at x→0 from the left
$$\lim_{x \to 0^+} \left(\frac{x - 2}{x}\right)^{2 x} = 1$$
More at x→0 from the right
$$\lim_{x \to 1^-} \left(\frac{x - 2}{x}\right)^{2 x} = 1$$
More at x→1 from the left
$$\lim_{x \to 1^+} \left(\frac{x - 2}{x}\right)^{2 x} = 1$$
More at x→1 from the right
$$\lim_{x \to -\infty} \left(\frac{x - 2}{x}\right)^{2 x} = e^{-4}$$
More at x→-oo
The graph