Mister Exam
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How to use it?
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Limit of (1+x^3-2*x)/(8+x^10-9*x)
- Limit of x^2+a/(x^3-a^3)-x*(1+a)
-
Identical expressions
- ((one +x)/(eleven +x))^x
- ((1 plus x) divide by (11 plus x)) to the power of x
- ((one plus x) divide by (eleven plus x)) to the power of x
- ((1+x)/(11+x))x
- 1+x/11+xx
- 1+x/11+x^x
- ((1+x) divide by (11+x))^x
-
Similar expressions
- ((1+x)/(11-x))^x
- ((1-x)/(11+x))^x
Limit of the function ((1+x)/(11+x))^x
The solution
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x
/1 + x \
lim |------|
x->oo\11 + x/
$$\lim_{x \to \infty} \left(\frac{x + 1}{x + 11}\right)^{x}$$
Limit(((1 + x)/(11 + x))^x, x, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{x \to \infty} \left(\frac{x + 1}{x + 11}\right)^{x}$$
transform
$$\lim_{x \to \infty} \left(\frac{x + 1}{x + 11}\right)^{x}$$
=
$$\lim_{x \to \infty} \left(\frac{\left(x + 11\right) - 10}{x + 11}\right)^{x}$$
=
$$\lim_{x \to \infty} \left(- \frac{10}{x + 11} + \frac{x + 11}{x + 11}\right)^{x}$$
=
$$\lim_{x \to \infty} \left(1 - \frac{10}{x + 11}\right)^{x}$$
=
do replacement
$$u = \frac{x + 11}{-10}$$
then
$$\lim_{x \to \infty} \left(1 - \frac{10}{x + 11}\right)^{x}$$ =
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- 10 u - 11}$$
=
$$\lim_{u \to \infty}\left(\frac{\left(1 + \frac{1}{u}\right)^{- 10 u}}{\left(1 + \frac{1}{u}\right)^{11}}\right)$$
=
$$\lim_{u \to \infty} \frac{1}{\left(1 + \frac{1}{u}\right)^{11}} \lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- 10 u}$$
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- 10 u}$$
=
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{-10}$$
The limit
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}$$
is second remarkable limit, is equal to e ~ 2.718281828459045
then
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{-10} = e^{-10}$$
The final answer:
$$\lim_{x \to \infty} \left(\frac{x + 1}{x + 11}\right)^{x} = e^{-10}$$
$$\lim_{x \to \infty} \left(\frac{x + 1}{x + 11}\right)^{x}$$
transform
$$\lim_{x \to \infty} \left(\frac{x + 1}{x + 11}\right)^{x}$$
=
$$\lim_{x \to \infty} \left(\frac{\left(x + 11\right) - 10}{x + 11}\right)^{x}$$
=
$$\lim_{x \to \infty} \left(- \frac{10}{x + 11} + \frac{x + 11}{x + 11}\right)^{x}$$
=
$$\lim_{x \to \infty} \left(1 - \frac{10}{x + 11}\right)^{x}$$
=
do replacement
$$u = \frac{x + 11}{-10}$$
then
$$\lim_{x \to \infty} \left(1 - \frac{10}{x + 11}\right)^{x}$$ =
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- 10 u - 11}$$
=
$$\lim_{u \to \infty}\left(\frac{\left(1 + \frac{1}{u}\right)^{- 10 u}}{\left(1 + \frac{1}{u}\right)^{11}}\right)$$
=
$$\lim_{u \to \infty} \frac{1}{\left(1 + \frac{1}{u}\right)^{11}} \lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- 10 u}$$
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- 10 u}$$
=
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{-10}$$
The limit
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}$$
is second remarkable limit, is equal to e ~ 2.718281828459045
then
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{-10} = e^{-10}$$
The final answer:
$$\lim_{x \to \infty} \left(\frac{x + 1}{x + 11}\right)^{x} = e^{-10}$$
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
$$\lim_{x \to \infty} \left(\frac{x + 1}{x + 11}\right)^{x} = e^{-10}$$
$$\lim_{x \to 0^-} \left(\frac{x + 1}{x + 11}\right)^{x} = 1$$
More at x→0 from the left
$$\lim_{x \to 0^+} \left(\frac{x + 1}{x + 11}\right)^{x} = 1$$
More at x→0 from the right
$$\lim_{x \to 1^-} \left(\frac{x + 1}{x + 11}\right)^{x} = \frac{1}{6}$$
More at x→1 from the left
$$\lim_{x \to 1^+} \left(\frac{x + 1}{x + 11}\right)^{x} = \frac{1}{6}$$
More at x→1 from the right
$$\lim_{x \to -\infty} \left(\frac{x + 1}{x + 11}\right)^{x} = e^{-10}$$
More at x→-oo
$$\lim_{x \to 0^-} \left(\frac{x + 1}{x + 11}\right)^{x} = 1$$
More at x→0 from the left
$$\lim_{x \to 0^+} \left(\frac{x + 1}{x + 11}\right)^{x} = 1$$
More at x→0 from the right
$$\lim_{x \to 1^-} \left(\frac{x + 1}{x + 11}\right)^{x} = \frac{1}{6}$$
More at x→1 from the left
$$\lim_{x \to 1^+} \left(\frac{x + 1}{x + 11}\right)^{x} = \frac{1}{6}$$
More at x→1 from the right
$$\lim_{x \to -\infty} \left(\frac{x + 1}{x + 11}\right)^{x} = e^{-10}$$
More at x→-oo
The graph