Mister Exam
Other calculators:
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How to use it?
- Limit of the function:
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Limit of (10+x^2-7*x)/(8-x^3)
-
Limit of (-3-5*x+2*x^2)/(-9+x^2)
-
Limit of (1+x^2)^(2/x)
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Limit of (1-3/x)^(2*x)
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Identical expressions
- (one - three /x)^(two *x)
- (1 minus 3 divide by x) to the power of (2 multiply by x)
- (one minus three divide by x) to the power of (two multiply by x)
- (1-3/x)(2*x)
- 1-3/x2*x
- (1-3/x)^(2x)
- (1-3/x)(2x)
- 1-3/x2x
- 1-3/x^2x
- (1-3 divide by x)^(2*x)
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Similar expressions
- (1+3/x)^(2*x)
Limit of the function (1-3/x)^(2*x)
The solution
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[src]
2*x
/ 3\
lim |1 - -|
x->oo\ x/
$$\lim_{x \to \infty} \left(1 - \frac{3}{x}\right)^{2 x}$$
Limit((1 - 3/x)^(2*x), x, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{x \to \infty} \left(1 - \frac{3}{x}\right)^{2 x}$$
transform
do replacement
$$u = \frac{x}{-3}$$
then
$$\lim_{x \to \infty} \left(1 - \frac{3}{x}\right)^{2 x}$$ =
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- 6 u}$$
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- 6 u}$$
=
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{-6}$$
The limit
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}$$
is second remarkable limit, is equal to e ~ 2.718281828459045
then
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{-6} = e^{-6}$$
The final answer:
$$\lim_{x \to \infty} \left(1 - \frac{3}{x}\right)^{2 x} = e^{-6}$$
$$\lim_{x \to \infty} \left(1 - \frac{3}{x}\right)^{2 x}$$
transform
do replacement
$$u = \frac{x}{-3}$$
then
$$\lim_{x \to \infty} \left(1 - \frac{3}{x}\right)^{2 x}$$ =
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- 6 u}$$
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- 6 u}$$
=
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{-6}$$
The limit
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}$$
is second remarkable limit, is equal to e ~ 2.718281828459045
then
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{-6} = e^{-6}$$
The final answer:
$$\lim_{x \to \infty} \left(1 - \frac{3}{x}\right)^{2 x} = e^{-6}$$
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
$$\lim_{x \to \infty} \left(1 - \frac{3}{x}\right)^{2 x} = e^{-6}$$
$$\lim_{x \to 0^-} \left(1 - \frac{3}{x}\right)^{2 x} = 1$$
More at x→0 from the left
$$\lim_{x \to 0^+} \left(1 - \frac{3}{x}\right)^{2 x} = 1$$
More at x→0 from the right
$$\lim_{x \to 1^-} \left(1 - \frac{3}{x}\right)^{2 x} = 4$$
More at x→1 from the left
$$\lim_{x \to 1^+} \left(1 - \frac{3}{x}\right)^{2 x} = 4$$
More at x→1 from the right
$$\lim_{x \to -\infty} \left(1 - \frac{3}{x}\right)^{2 x} = e^{-6}$$
More at x→-oo
$$\lim_{x \to 0^-} \left(1 - \frac{3}{x}\right)^{2 x} = 1$$
More at x→0 from the left
$$\lim_{x \to 0^+} \left(1 - \frac{3}{x}\right)^{2 x} = 1$$
More at x→0 from the right
$$\lim_{x \to 1^-} \left(1 - \frac{3}{x}\right)^{2 x} = 4$$
More at x→1 from the left
$$\lim_{x \to 1^+} \left(1 - \frac{3}{x}\right)^{2 x} = 4$$
More at x→1 from the right
$$\lim_{x \to -\infty} \left(1 - \frac{3}{x}\right)^{2 x} = e^{-6}$$
More at x→-oo
The graph