Limits of Indeterminate Forms: lim[x->infty] (1+5/x+4/x^2)^x

[Klementine]

Hi there, I was given these problems to solve:

1. Solve as x approaches infinity the following: (1+5/x+4/x^2)^x
I understand the answer is e^5 and I have been able to start the problem, putting ln on both sides and then dividing by "x^-1" but I just can't quite find the real answer.

2. Solve as x approaches 0+ the following: (cos(2x))^(3/x^2)
Again, I'm stuck with some long complicated answer that does not make sense. I do not know the real answer for this one.

3. Solve as x approaches 0+ the following: (5x)^x^9
This one is the one on which I am the most stumped. How on earth does one find the derivative of (5x)^x?

Thank you!

 

[Deleted member 4993]

Hi there, I was given these problems to solve:

1. Solve as x approaches infinity the following: (1+5/x+4/x^2)^x
I understand the answer is e^5 and I have been able to start the problem, putting ln on both sides and then dividing by "x^-1" but I just can't quite find the real answer.

2. Solve as x approaches 0+ the following: (cos(2x))^(3/x^2)
Again, I'm stuck with some long complicated answer that does not make sense. I do not know the real answer for this one.

3. Solve as x approaches 0+ the following: (5x)^x^9
This one is the one on which I am the most stumped. How on earth does one find the derivative of (5x)^x?

Thank you!

3. Solve as x approaches 0+ the following: (5x)^x^9
This one is the one on which I am the most stumped. How on earth does one find the derivative of (5x)^x?

With divine inspiration - use logarithm.

y = x^x

ln(y) = x * ln(x)

d/dx[ln(y)] = d/dx[x * ln(x)]

1/y * dy/dx = ln(x) + 1

and continue.....

 

[stapel]

\(\displaystyle \mbox{1. Solve }\, \)\(\displaystyle \displaystyle \lim_{x\, \rightarrow\, \infty}\, \)\(\displaystyle \left(1\, +\, \dfrac{5}{x}\, +\, \dfrac{4}{x^2}\right)^x\)

I understand the answer is e^5 and I have been able to start the problem, putting ln on both sides and then dividing by "x^-1" but I just can't quite find the real answer.

When you say "putting ln on both sides", what is the other "side"? Do you understand that this is "taking the log of each side" of some equation? (What equation?) Are you saying that you started somewhat along the lines of the following?

. . .The original exercise:

. . . . .\(\displaystyle \displaystyle \lim_{x\, \rightarrow\, \infty}\, \)\(\displaystyle \left(1\, +\, \dfrac{5}{x}\, +\, \dfrac{4}{x^2}\right)^x\)

. . .Consider just the expression, sans limit, and name it:

. . . . .\(\displaystyle y\, =\, \left(1\, +\, \dfrac{5}{x}\, +\, \dfrac{4}{x^2}\right)^x\)

. . .Take the natural log of each side:

. . . . .\(\displaystyle \ln(y)\, =\, \ln\bigg(\left(1\, +\, \dfrac{5}{x}\, +\, \dfrac{4}{x^2}\right)^x\bigg)\)

. . .Apply a log rule:

. . . . .\(\displaystyle \ln(y)\, =\, x\, \ln\left(1\, +\, \dfrac{5}{x}\, +\, \dfrac{4}{x^2}\right)\)

. . .Work with the fractions:

. . . . .\(\displaystyle \ln(y)\, =\, x\, \ln\left(\dfrac{x^2\, +\, 5x\, +\, 4}{x^2}\right)\)

. . .Apply another log rule:

. . . . .\(\displaystyle \ln(y)\, =\, x\, \bigg(\ln(x^2\, +\, 5x\, +\, 4)\, -\, \ln(x^2)\bigg)\)

What were your next steps? How did you arrive at "dividing by x^-1" (which is the same thing as multiplying by x)?

\(\displaystyle \mbox{2. Solve }\, \)\(\displaystyle \displaystyle \lim_{x\, \rightarrow\, 0^+}\, \)\(\displaystyle \left(\cos(2x)\right)^{3/x^2}\)

Again, I'm stuck with some long complicated answer that does not make sense. I do not know the real answer for this one.

What is your answer? How did you get there? What were your steps?

\(\displaystyle \mbox{3. Solve }\, \)\(\displaystyle \displaystyle \lim_{x\, \rightarrow\, 0^+}\, \)\(\displaystyle {(5x)^x}^9\)

This one is the one on which I am the most stumped. How on earth does one find the derivative of (5x)^x?

Why did you take logs on the first exercise? What is different (or the same) here?

When you reply, please show your steps and reasoning, as I did above, so we can see what's going on. Thank you!