Mathematics, 29.07.2021 23:40 juliawatakip5fmg7

Lim ₓ→∞ (x+4/x-1)∧x+4

Answers: 3

Show answers

Answers

Answer from: AutumnJoy12

It looks like the limit you want to find is

$\displaystyle \lim_{x\to\infty} \left(\frac{x+4}{x-1}\right)^{x+4}$

One way to compute this limit relies only on the definition of the constant e and some basic properties of limits. In particular,

$e = \displaystyle\lim_{x\to\infty}\left(1+\frac1x\right)^x$

The idea is to recast the given limit to make it resemble this definition. The definition contains a fraction with x as its denominator. If we expand the fraction in the given limand, we have a denominator of x - 1. So we rewrite everything in terms of x - 1 :

$\left(\dfrac{x+4}{x-1}\right)^{x+4} = \left(\dfrac{x-1+5}{x-1}\right)^{x-1+5} \\\\ = \left(1+\dfrac5{x-1}\right)^{x-1+5} \\\\ =\left(1+\dfrac5{x-1}\right)^{x-1} \times \left(1+\dfrac5{x-1}\right)^5$

Now in the first term of this product, we substitute y = (x - 1)/5 :

$\left(\dfrac{x+4}{x-1}\right)^{x+4} = \left(1+\dfrac1y\right)^{5y} \times \left(1+\dfrac5{x-1}\right)^5$

Then use a property of exponentiation to write this as

$\left(\dfrac{x+4}{x-1}\right)^{x+4} = \left(\left(1+\dfrac1y\right)^y\right)^5 \times \left(1+\dfrac5{x-1}\right)^5$

In terms of end behavior, (x - 1)/5 and x behave the same way because they both approach ∞ at a proportional rate, so we can essentially y with x. Then by applying some limit properties, we have

$\displaystyle \lim_{x\to\infty} \left(\frac{x+4}{x-1}\right)^{x+4} = \lim_{x\to\infty} \left(\left(1+\dfrac1x\right)^x\right)^5 \times \left(1+\dfrac5{x-1}\right)^5 \\\\ = \lim_{x\to\infty}\left(\left(1+\dfrac1x\right)^x\right)^5 \times \lim_{x\to\infty}\left(1+\dfrac5{x-1}\right)^5 \\\\ =\left(\lim_{x\to\infty}\left(1+\dfrac1x\right)^x\right)^5 \times \left(\lim_{x\to\infty}\left(1+\dfrac5{x-1}\right)\right)^5$

By definition, the first limit is e and the second limit is 1, so that

$\displaystyle \lim_{x\to\infty} \left(\frac{x+4}{x-1}\right)^{x+4} = e^5\times1^5 = \boxed{e^5}$

You can also use L'Hopital's rule to compute it. Evaluating the limit "directly" at infinity results in the indeterminate form $1^\infty$ .

Rewrite

$\left(\dfrac{x+4}{x-1}\right)^{x+4} = \exp\left((x+4)\ln\dfrac{x+4}{x-1}\right)$

so that

$\displaystyle \lim_{x\to\infty} \left(\frac{x+4}{x-1}\right)^{x+4} = \lim_{x\to\infty}\exp\left((x+4)\ln\dfrac{x+4}{x-1}\right) \\\\ = \exp\left(\lim_{x\to\infty}(x+4)\ln\dfrac{x+4}{x-1}\right) \\\\ =\exp\left(\lim_{x\to\infty}\frac{\ln\dfrac{x+4}{x-1}}{\dfrac1{x+4}}\right)$

and now evaluating "directly" at infinity gives the indeterminate form 0/0, making the limit ready for L'Hopital's rule.

We have

$\dfrac{\mathrm d}{\mathrm dx}\left[\ln\dfrac{x+4}{x-1}\right] = -\dfrac5{(x-1)^2}\times\dfrac{1}{\frac{x+4}{x-1}} = -\dfrac5{(x-1)(x+4)}$

$\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1{x+4}\right]=-\dfrac1{(x+4)^2}$

and so

$\displaystyle \exp\left(\lim_{x\to\infty}\frac{\ln\dfrac{x+4}{x-1}}{\dfrac1{x+4}}\right) = \exp\left(\lim_{x\to\infty}\frac{-\dfrac5{(x-1)(x+4)}}{-\dfrac1{(x+4)^2}}\right) \\\\ = \exp\left(5\lim_{x\to\infty}\frac{x+4}{x-1}\right) \\\\ = \exp(5) = \boxed{e^5}$

Answer from: Quest

Intersecting secant-tangent theorem states that if a tangent segment and a secant segment are drawn to a circle from an exterior point, then the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment.so to break it down: 1. pq acts as the tangent and ps as the secant with an intersection at p.sr = 21rp = 3x + 3pq = 4x + 4following to equation: tangent squared = the two segments multiplied by each otherso (4x + 4) ^ 2 = (3x + 3) • (21) simplifies to 16x^2 - 31x - 47 = 0 or (16x - 47)(x + 1) x = -1 or x = 47/162. pq acts as the tangent and ps as the secant with an intersection at p (again! )sr = 16rp = xpq = 15(15) ^ 2 = 16 • x225 = 16xx = 225/163. the interesting chords are proportional to each other so a ratio is possible to set up: ap dp 10 3 + x = = —— = pc pb 8 xcross multiple to get 10x = (3 + x)(8)x = 12

Answer from: Quest

answer: both chords and diameters have two endpoints on a circle. diameters must intersect the center of a circle.

Another question on Mathematics

Mathematics, 21.06.2019 14:00

Arandom number generator is used to model the patterns of animals in the wild this type of study is called

Answers: 1

Answer

Mathematics, 21.06.2019 17:00

Asap i need it now choose all the answers that apply. sex-linked disorders affect males more than females affect females more than males can be carried by females, without being expressed are always expressed in males are caused by genes carried on the x and y chromosomes

Answers: 1

Answer

Mathematics, 21.06.2019 18:00

Does anyone know how to do geometryme with this problem

Answers: 1

Answer

Mathematics, 22.06.2019 00:30

What is the sum of the geometric series in which a1 = 7, r = 3, and an = 1,701? hint: cap s sub n equals start fraction a sub one left parenthesis one minus r to the power of n end power right parenthesis over one minus r end fraction comma r ≠ 1, where a1 is the first term and r is the common ratio

Answers: 1

Answer

You know the right answer?

Lim ₓ→∞ (x+4/x-1)∧x+4...

Questions

Mathematics, 05.07.2019 10:30

[tex] \sqrt{2x + 3} + \sqrt{x - 7} = 3[/tex]what is x? i don't remember how to solve this type of problem so an explanation would be highly appreciat...

Mathematics, 05.07.2019 10:30

(do it if you want the original height of khafre's pyramid, next to the great pyramid, was about 471 ft. each side of its square base was about...

Mathematics, 05.07.2019 10:30

Which identifies they y intercept slope and zero of the linear function 3x-8y=24...