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