Mister Exam
Other calculators:
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How to use it?
- Limit of the function:
-
Limit of (15-28*x^5-19*x+16*x^2+19*x^3)/(15-30*x^2-7*x^5+11*x^4+23*x)
-
Limit of (6-30*x^5-24*x^6-15*x-8*x^2+15*x^4)/(22-7*x^2+5*x^6+10*x^4+26*x^3+26*x^5+27*x)
- Limit of (1-x)*tan(p*x/2)
-
Limit of 2*(1-x)*log(x)
-
Identical expressions
- ((six +x)/(four +x))^(one +x)
- ((6 plus x) divide by (4 plus x)) to the power of (1 plus x)
- ((six plus x) divide by (four plus x)) to the power of (one plus x)
- ((6+x)/(4+x))(1+x)
- 6+x/4+x1+x
- 6+x/4+x^1+x
- ((6+x) divide by (4+x))^(1+x)
-
Similar expressions
- ((6-x)/(4+x))^(1+x)
- ((6+x)/(4+x))^(1-x)
- ((6+x)/(4-x))^(1+x)
Limit of the function ((6+x)/(4+x))^(1+x)
The solution
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1 + x
/6 + x\
lim |-----|
x->oo\4 + x/
$$\lim_{x \to \infty} \left(\frac{x + 6}{x + 4}\right)^{x + 1}$$
Limit(((6 + x)/(4 + x))^(1 + x), x, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{x \to \infty} \left(\frac{x + 6}{x + 4}\right)^{x + 1}$$
transform
$$\lim_{x \to \infty} \left(\frac{x + 6}{x + 4}\right)^{x + 1}$$
=
$$\lim_{x \to \infty} \left(\frac{\left(x + 4\right) + 2}{x + 4}\right)^{x + 1}$$
=
$$\lim_{x \to \infty} \left(\frac{x + 4}{x + 4} + \frac{2}{x + 4}\right)^{x + 1}$$
=
$$\lim_{x \to \infty} \left(1 + \frac{2}{x + 4}\right)^{x + 1}$$
=
do replacement
$$u = \frac{x + 4}{2}$$
then
$$\lim_{x \to \infty} \left(1 + \frac{2}{x + 4}\right)^{x + 1}$$ =
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{2 u - 3}$$
=
$$\lim_{u \to \infty}\left(\frac{\left(1 + \frac{1}{u}\right)^{2 u}}{\left(1 + \frac{1}{u}\right)^{3}}\right)$$
=
$$\lim_{u \to \infty} \frac{1}{\left(1 + \frac{1}{u}\right)^{3}} \lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{2 u}$$
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{2 u}$$
=
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{2}$$
The limit
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}$$
is second remarkable limit, is equal to e ~ 2.718281828459045
then
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{2} = e^{2}$$
The final answer:
$$\lim_{x \to \infty} \left(\frac{x + 6}{x + 4}\right)^{x + 1} = e^{2}$$
$$\lim_{x \to \infty} \left(\frac{x + 6}{x + 4}\right)^{x + 1}$$
transform
$$\lim_{x \to \infty} \left(\frac{x + 6}{x + 4}\right)^{x + 1}$$
=
$$\lim_{x \to \infty} \left(\frac{\left(x + 4\right) + 2}{x + 4}\right)^{x + 1}$$
=
$$\lim_{x \to \infty} \left(\frac{x + 4}{x + 4} + \frac{2}{x + 4}\right)^{x + 1}$$
=
$$\lim_{x \to \infty} \left(1 + \frac{2}{x + 4}\right)^{x + 1}$$
=
do replacement
$$u = \frac{x + 4}{2}$$
then
$$\lim_{x \to \infty} \left(1 + \frac{2}{x + 4}\right)^{x + 1}$$ =
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{2 u - 3}$$
=
$$\lim_{u \to \infty}\left(\frac{\left(1 + \frac{1}{u}\right)^{2 u}}{\left(1 + \frac{1}{u}\right)^{3}}\right)$$
=
$$\lim_{u \to \infty} \frac{1}{\left(1 + \frac{1}{u}\right)^{3}} \lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{2 u}$$
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{2 u}$$
=
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{2}$$
The limit
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}$$
is second remarkable limit, is equal to e ~ 2.718281828459045
then
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{2} = e^{2}$$
The final answer:
$$\lim_{x \to \infty} \left(\frac{x + 6}{x + 4}\right)^{x + 1} = e^{2}$$
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
$$\lim_{x \to \infty} \left(\frac{x + 6}{x + 4}\right)^{x + 1} = e^{2}$$
$$\lim_{x \to 0^-} \left(\frac{x + 6}{x + 4}\right)^{x + 1} = \frac{3}{2}$$
More at x→0 from the left
$$\lim_{x \to 0^+} \left(\frac{x + 6}{x + 4}\right)^{x + 1} = \frac{3}{2}$$
More at x→0 from the right
$$\lim_{x \to 1^-} \left(\frac{x + 6}{x + 4}\right)^{x + 1} = \frac{49}{25}$$
More at x→1 from the left
$$\lim_{x \to 1^+} \left(\frac{x + 6}{x + 4}\right)^{x + 1} = \frac{49}{25}$$
More at x→1 from the right
$$\lim_{x \to -\infty} \left(\frac{x + 6}{x + 4}\right)^{x + 1} = e^{2}$$
More at x→-oo
$$\lim_{x \to 0^-} \left(\frac{x + 6}{x + 4}\right)^{x + 1} = \frac{3}{2}$$
More at x→0 from the left
$$\lim_{x \to 0^+} \left(\frac{x + 6}{x + 4}\right)^{x + 1} = \frac{3}{2}$$
More at x→0 from the right
$$\lim_{x \to 1^-} \left(\frac{x + 6}{x + 4}\right)^{x + 1} = \frac{49}{25}$$
More at x→1 from the left
$$\lim_{x \to 1^+} \left(\frac{x + 6}{x + 4}\right)^{x + 1} = \frac{49}{25}$$
More at x→1 from the right
$$\lim_{x \to -\infty} \left(\frac{x + 6}{x + 4}\right)^{x + 1} = e^{2}$$
More at x→-oo
The graph