Mister Exam
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How to use it?
- Limit of the function:
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Limit of ((-2+3*x)/(1+3*x))^(2*x)
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Limit of ((-4+3*x)/(2+3*x))^(2*x)
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Limit of (1-cos(4*x))/(1-cos(8*x))
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Limit of (1+x^2-4*x)/(1+2*x)
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Identical expressions
- ((- four + three *x)/(two + three *x))^(two *x)
- (( minus 4 plus 3 multiply by x) divide by (2 plus 3 multiply by x)) to the power of (2 multiply by x)
- (( minus four plus three multiply by x) divide by (two plus three multiply by x)) to the power of (two multiply by x)
- ((-4+3*x)/(2+3*x))(2*x)
- -4+3*x/2+3*x2*x
- ((-4+3x)/(2+3x))^(2x)
- ((-4+3x)/(2+3x))(2x)
- -4+3x/2+3x2x
- -4+3x/2+3x^2x
- ((-4+3*x) divide by (2+3*x))^(2*x)
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Similar expressions
- ((4+3*x)/(2+3*x))^(2*x)
- ((-4+3*x)/(2-3*x))^(2*x)
- ((-4-3*x)/(2+3*x))^(2*x)
Limit of the function ((-4+3*x)/(2+3*x))^(2*x)
The solution
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2*x
/-4 + 3*x\
lim |--------|
x->oo\2 + 3*x /
$$\lim_{x \to \infty} \left(\frac{3 x - 4}{3 x + 2}\right)^{2 x}$$
Limit(((-4 + 3*x)/(2 + 3*x))^(2*x), x, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{x \to \infty} \left(\frac{3 x - 4}{3 x + 2}\right)^{2 x}$$
transform
$$\lim_{x \to \infty} \left(\frac{3 x - 4}{3 x + 2}\right)^{2 x}$$
=
$$\lim_{x \to \infty} \left(\frac{\left(3 x + 2\right) - 6}{3 x + 2}\right)^{2 x}$$
=
$$\lim_{x \to \infty} \left(- \frac{6}{3 x + 2} + \frac{3 x + 2}{3 x + 2}\right)^{2 x}$$
=
$$\lim_{x \to \infty} \left(1 - \frac{6}{3 x + 2}\right)^{2 x}$$
=
do replacement
$$u = \frac{3 x + 2}{-6}$$
then
$$\lim_{x \to \infty} \left(1 - \frac{6}{3 x + 2}\right)^{2 x}$$ =
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- 4 u - \frac{4}{3}}$$
=
$$\lim_{u \to \infty}\left(\frac{\left(1 + \frac{1}{u}\right)^{- 4 u}}{\left(1 + \frac{1}{u}\right)^{\frac{4}{3}}}\right)$$
=
$$\lim_{u \to \infty} \frac{1}{\left(1 + \frac{1}{u}\right)^{\frac{4}{3}}} \lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- 4 u}$$
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- 4 u}$$
=
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{-4}$$
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:
$$\lim_{x \to \infty} \left(\frac{3 x - 4}{3 x + 2}\right)^{2 x} = e^{-4}$$
$$\lim_{x \to \infty} \left(\frac{3 x - 4}{3 x + 2}\right)^{2 x}$$
transform
$$\lim_{x \to \infty} \left(\frac{3 x - 4}{3 x + 2}\right)^{2 x}$$
=
$$\lim_{x \to \infty} \left(\frac{\left(3 x + 2\right) - 6}{3 x + 2}\right)^{2 x}$$
=
$$\lim_{x \to \infty} \left(- \frac{6}{3 x + 2} + \frac{3 x + 2}{3 x + 2}\right)^{2 x}$$
=
$$\lim_{x \to \infty} \left(1 - \frac{6}{3 x + 2}\right)^{2 x}$$
=
do replacement
$$u = \frac{3 x + 2}{-6}$$
then
$$\lim_{x \to \infty} \left(1 - \frac{6}{3 x + 2}\right)^{2 x}$$ =
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- 4 u - \frac{4}{3}}$$
=
$$\lim_{u \to \infty}\left(\frac{\left(1 + \frac{1}{u}\right)^{- 4 u}}{\left(1 + \frac{1}{u}\right)^{\frac{4}{3}}}\right)$$
=
$$\lim_{u \to \infty} \frac{1}{\left(1 + \frac{1}{u}\right)^{\frac{4}{3}}} \lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- 4 u}$$
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- 4 u}$$
=
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{-4}$$
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:
$$\lim_{x \to \infty} \left(\frac{3 x - 4}{3 x + 2}\right)^{2 x} = e^{-4}$$
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
The graph