# [SOLVED]Another two limits at infinity

#### wishmaster

##### Active member
I know im already boring with limits,but i again have two of them to deal with and i dont know how.....

1.
$$\displaystyle \lim _{n \to \infty} \frac{5^{n^+1}-2*5^n+5^{n-1}}{3^{n+1}-3^n}$$

2.
$$\displaystyle \lim _{n \to \infty} \frac{\sqrt[3]{n^4}+\sqrt{n}+1}{\sqrt[6]{n^4}+\sqrt[3]{n}+2}$$

#### Chris L T521

##### Well-known member
Staff member
I know im already boring with limits,but i again have two of them to deal with and i dont know how.....

1.
$$\displaystyle \lim _{n \to \infty} \frac{5^{n^+1}-2*5^n+5^{n-1}}{3^{n+1}-3^n}$$
Divide each term in the limit by $3^{n+1}$. It should become more apparent as to what the limit is once you do this.

2.
$$\displaystyle \lim _{n \to \infty} \frac{\sqrt[3]{n^4}+\sqrt{n}+1}{\sqrt[6]{n^4}+\sqrt[3]{n}+2}$$
Similarly , divide each term in the limit by $\sqrt[6]{n^4}$ and then evaluate the limit. What do you get when you do this?

#### MarkFL

Staff member
1.) By factoring, you can get this limit to the form:

$$\displaystyle L=a\cdot\lim_{n\to\infty}b^n$$

where $0<a,b\in\mathbb{R}$.

At this point, you may divide through by $a$ to obtain:

$$\displaystyle \frac{L}{a}=\lim_{n\to\infty}b^n$$

Next, take the natural log of both sides, and apply the property of limits:

$$\displaystyle \log_a\left(\lim_{x\to c}f(x) \right)=\lim_{x\to c}\left(\log_a\left(f(x) \right) \right)$$

to obtain:

$$\displaystyle \ln\left(\frac{L}{a} \right)=\lim_{n\to\infty}\ln\left(b^n \right)$$

Using the log property $$\displaystyle \log_a\left(b^c \right)=c\cdot\log_a(b)$$ we may write:

$$\displaystyle \ln\left(\frac{L}{a} \right)=\lim_{n\to\infty}n\cdot\ln\left(b \right)$$

$$\displaystyle \ln\left(\frac{L}{a} \right)=\ln\left(b \right)\lim_{n\to\infty}n$$

If $0<b<1$, we have:

$$\displaystyle \ln\left(\frac{L}{a} \right)=-\infty$$

Converting from logarithmic to exponential form, we have:

$$\displaystyle \frac{L}{a}=e^{-\infty}=0\implies L=0$$

If $b=1$ then we have:

$$\displaystyle \ln\left(\frac{L}{a} \right)=0$$

Converting from logarithmic to exponential form:

$$\displaystyle \frac{L}{a}=1\implies L=a$$

If $1<b$ then we have:

$$\displaystyle \ln\left(\frac{L}{a} \right)=\infty$$

Converting from logarithmic to exponential form, we find:

$$\displaystyle \frac{L}{a}=e^{\infty}=\infty\implies L=\infty$$

What do you find?

#### wishmaster

##### Active member
1.) By factoring, you can get this limit to the form:

$$\displaystyle L=a\cdot\lim_{n\to\infty}b^n$$

where $0<a,b\in\mathbb{R}$.

At this point, you may divide through by $a$ to obtain:

$$\displaystyle \frac{L}{a}=\lim_{n\to\infty}b^n$$

Next, take the natural log of both sides, and apply the property of limits:

$$\displaystyle \log_a\left(\lim_{x\to c}f(x) \right)=\lim_{x\to c}\left(\log_a\left(f(x) \right) \right)$$

to obtain:

$$\displaystyle \ln\left(\frac{L}{a} \right)=\lim_{n\to\infty}\ln\left(b^n \right)$$

Using the log property $$\displaystyle \log_a\left(b^c \right)=c\cdot\log_a(b)$$ we may write:

$$\displaystyle \ln\left(\frac{L}{a} \right)=\lim_{n\to\infty}n\cdot\ln\left(b \right)$$

$$\displaystyle \ln\left(\frac{L}{a} \right)=\ln\left(b \right)\lim_{n\to\infty}n$$

If $0<b<1$, we have:

$$\displaystyle \ln\left(\frac{L}{a} \right)=-\infty$$

Converting from logarithmic to exponential form, we have:

$$\displaystyle \frac{L}{a}=e^{-\infty}=0\implies L=0$$

If $b=1$ then we have:

$$\displaystyle \ln\left(\frac{L}{a} \right)=0$$

Converting from logarithmic to exponential form:

$$\displaystyle \frac{L}{a}=1\implies L=a$$

If $1<b$ then we have:

$$\displaystyle \ln\left(\frac{L}{a} \right)=\infty$$

Converting from logarithmic to exponential form, we find:

$$\displaystyle \frac{L}{a}=e^{\infty}=\infty\implies L=\infty$$

What do you find?
Thats gonna be a hard one,i think im not able to solve this.....

#### MarkFL

Staff member
Thats gonna be a hard one,i think im not able to solve this.....
You really don't need to use the logarithmic approach I gave. I wrote it out to demonstrate how the value of $b$ affects the outcome of the limit.

Consider these two limits:

$$\displaystyle \lim_{n\to\infty}\left(\frac{2}{3} \right)^n$$

$$\displaystyle \lim_{n\to\infty}\left(\frac{3}{2} \right)^n$$

How would you respond to these?

#### wishmaster

##### Active member
You really don't need to use the logarithmic approach I gave. I wrote it out to demonstrate how the value of $b$ affects the outcome of the limit.

Consider these two limits:

$$\displaystyle \lim_{n\to\infty}\left(\frac{2}{3} \right)^n$$

$$\displaystyle \lim_{n\to\infty}\left(\frac{3}{2} \right)^n$$

How would you respond to these?
They both go to infinity?

#### MarkFL

Staff member
They both go to infinity?
Well, one of them does. What happens if we multiply a positive number by $\dfrac{2}{3}$ (or any positive real number less than 1)? Does it get larger or smaller?

#### wishmaster

##### Active member
Well, one of them does. What happens if we multiply a positive number by $\dfrac{2}{3}$ (or any positive real number less than 1)? Does it get larger or smaller?
Smaller.

So $$\displaystyle \frac{2}{3}$$ is going to zero,while $$\displaystyle \frac{3}{2}$$ is going to infinity

#### Chris L T521

##### Well-known member
Staff member
Smaller.

So $$\displaystyle \frac{2}{3}$$ is going to zero,while $$\displaystyle \frac{3}{2}$$ is going to infinity
Well....more technically $\left(\dfrac{2}{3}\right)^n$ goes to zero whereas $\left(\dfrac{3}{2}\right)^n$ goes to $\infty$. With this known, what can you conclude about your original limit? If you follow the hint I provided, you'll see that you can reduce the limit to $\displaystyle \lim_{n\to\infty}\frac{8}{5}\left( \frac{5}{3}\right)^n$.

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#### MarkFL

Staff member
Smaller.

So $$\displaystyle \frac{2}{3}$$ is going to zero,while $$\displaystyle \frac{3}{2}$$ is going to infinity
Yes...but see the post above by Chris L T521 for clarification.

If we write the expression as:

$$\displaystyle \left(\frac{2}{3} \right)^n=\frac{2^n}{3^n}$$

And then observe that this ratio gets smaller and smaller as $n$ grows without bound, we then conclude that the limit goes to zero.

Likewise if we have:

$$\displaystyle \left(\frac{3}{2} \right)^n=\frac{3^n}{2^n}$$

And then observe that this ratio gets larger and larger as $n$ grows without bound, we then conclude that the limit goes to infinity.

#### wishmaster

##### Active member
Yes...but see the post above by Chris L T521 for clarification.

If we write the expression as:

$$\displaystyle \left(\frac{2}{3} \right)^n=\frac{2^n}{3^n}$$

And then observe that this ratio gets smaller and smaller as $n$ grows without bound, we then conclude that the limit goes to zero.

Likewise if we have:

$$\displaystyle \left(\frac{3}{2} \right)^n=\frac{3^n}{2^n}$$

And then observe that this ratio gets larger and larger as $n$ grows without bound, we then conclude that the limit goes to infinity.

i understand that....question is how to solve limits

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
May I ask the OP what course is he taking that discusses these limits ?

#### MarkFL

Staff member
...If you follow the hint I provided, you'll see that you can reduce the limit to $\displaystyle \lim_{n\to\infty}\frac{16}{15}\left( \frac{5}{3}\right)^n$.
I get a slightly different result:

$$\displaystyle L=\lim _{n\to\infty} \frac{5^{n+1}-2\cdot5^n+5^{n-1}}{3^{n+1}-3^n}$$

$$\displaystyle L=\lim _{n\to\infty} \frac{5^{n-1}\left(5^{2}-2\cdot5+1 \right)}{3^n(3-1)}$$

$$\displaystyle L=\lim _{n\to\infty} \frac{16\cdot5^{n-1}}{2\cdot3^n}$$

$$\displaystyle L=\lim _{n\to\infty} \frac{8\cdot5^n}{5\cdot3^n}$$

$$\displaystyle L=\frac{8}{5}\lim _{n\to\infty} \left(\frac{5}{3} \right)^n$$

#### MarkFL

Staff member
i understand that....question is how to solve limits
Understanding this is key to evaluating the given limit.

#### wishmaster

##### Active member
I get a slightly different result:

$$\displaystyle L=\lim _{n\to\infty} \frac{5^{n+1}-2\cdot5^n+5^{n-1}}{3^{n+1}-3^n}$$

$$\displaystyle L=\lim _{n\to\infty} \frac{5^{n-1}\left(5^{2}-2\cdot5+1 \right)}{3^n(3-1)}$$

$$\displaystyle L=\lim _{n\to\infty} \frac{16\cdot5^{n-1}}{2\cdot3^n}$$

$$\displaystyle L=\lim _{n\to\infty} \frac{8\cdot5^n}{5\cdot3^n}$$

$$\displaystyle L=\frac{8}{5}\lim _{n\to\infty} \left(\frac{5}{3} \right)^n$$

So in numerator you have exposed $$\displaystyle 5^{n-1}$$ and in denominator $$\displaystyle 3^n$$?

#### MarkFL

Staff member
So in numerator you have exposed $$\displaystyle 5^{n-1}$$ and in denominator $$\displaystyle 3^n$$?
Yes, you want to expose (or factor out) the smallest power, because it is a common factor to all terms.

#### wishmaster

##### Active member
Yes, you want to expose (or factor out) the smallest power, because it is a common factor to all terms.
Yes,i understand now.....

Can you show me for the other limit?

#### Chris L T521

##### Well-known member
Staff member
I get a slightly different result:

$$\displaystyle L=\lim _{n\to\infty} \frac{5^{n+1}-2\cdot5^n+5^{n-1}}{3^{n+1}-3^n}$$

$$\displaystyle L=\lim _{n\to\infty} \frac{5^{n-1}\left(5^{2}-2\cdot5+1 \right)}{3^n(3-1)}$$

$$\displaystyle L=\lim _{n\to\infty} \frac{16\cdot5^{n-1}}{2\cdot3^n}$$

$$\displaystyle L=\lim _{n\to\infty} \frac{8\cdot5^n}{5\cdot3^n}$$

$$\displaystyle L=\frac{8}{5}\lim _{n\to\infty} \left(\frac{5}{3} \right)^n$$
Woops, I misread my own handwriting...

I got the same thing you did. XD

#### MarkFL

Staff member
Yes,i understand now.....

Can you show me for the other limit?
Chris L T521 gave you an excellent suggestion for evaluating the second limit in post #2.

Convert the terms from radical to rational exponent notation:

$$\displaystyle \sqrt[n]{a^m}=a^{\frac{m}{n}}$$

and then use the property of exponents:

$$\displaystyle \frac{a^b}{a^c}=a^{b-c}$$

when doing the suggested divisions. What do you get?

#### wishmaster

##### Active member
Chris L T521 gave you an excellent suggestion for evaluating the second limit in post #2.

Convert the terms from radical to rational exponent notation:

$$\displaystyle \sqrt[n]{a^m}=a^{\frac{m}{n}}$$

and then use the property of exponents:

$$\displaystyle \frac{a^b}{a^c}=a^{b-c}$$

when doing the suggested divisions. What do you get?
$$\displaystyle \frac{n^{\frac{4}{3}}+n^{\frac{1}{2}}+1}{n^{\frac{4}{6}}+n^{\frac{2}{3}}+2}$$

That equals to: $$\displaystyle n^{\frac{4}{6}}-n^{\frac{1}{6}}+3$$

Am i right?

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#### MarkFL

Staff member
$$\displaystyle \frac{n^{\frac{4}{3}}+n^{\frac{1}{2}}+1}{n^{\frac{4}{6}}+n^{\frac{2}{3}}+2}$$
That's not quite right...look at the second term in the denominator again.

Once you fix this, then do the suggested division.

#### wishmaster

##### Active member
That's not quite right...look at the second term in the denominator again.

Once you fix this, then do the suggested division.
Second term should be $$\displaystyle n^{\frac{1}{3}}$$ ??

#### wishmaster

##### Active member
Second term should be $$\displaystyle n^{\frac{1}{3}}$$ ??
Then is $$\displaystyle \frac{n^{\frac{3}{2}}+1}{n+2}$$ ??

#### MarkFL

Staff member
Second term should be $$\displaystyle n^{\frac{1}{3}}$$ ??
Correct:

$$\displaystyle \sqrt[3]{n}=\sqrt[3]{n^1}=n^{\frac{1}{3}}$$

So, what do you get when you divide each term by:

$$\displaystyle n^{\frac{4}{6}}=n^{\frac{2}{3}}$$ ?

#### MarkFL

Then is $$\displaystyle \frac{n^{\frac{3}{2}}+1}{n+2}$$ ??