Another two limits at infinity

[theakdad]

I know I am already boring with limits,but i again have two of them to deal with and i don't 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]

I know I am already boring with limits,but i again have two of them to deal with and i don't 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]

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?

 

[theakdad]

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 going to be a hard one,i think I am not able to solve this...

 

[MarkFL]

Thats going to be a hard one,i think I am 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?

 

[theakdad]

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]

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?

 

[theakdad]

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]

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 $\displaystyle \lim_{n\to\infty}\frac{8}{5}\left( \frac{5}{3}\right)^n$.

 

[MarkFL]

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. :D

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.

 

[theakdad]

Yes...but see the post above by Chris L T521 for clarification. :D

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 ;)

 

[alyafey22]

May I ask the OP what course is he taking that discusses these limits ?

 

[MarkFL]

...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]

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

Understanding this is key to evaluating the given limit.

 

[theakdad]

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]

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.

 

[theakdad]

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]

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... (Headbang)

I got the same thing you did. XD

 

[MarkFL]

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?

 

[theakdad]

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?

 

[MarkFL]

\(\displaystyle \frac{n^{\frac{4}{3}}+n^{\frac{1}{2}}+1}{n^{\frac{4}{6}}+n^{\frac{2}{3}}+2} \)

That's not quite right...look at the second term in the denominator again. :D

Once you fix this, then do the suggested division.

 

[theakdad]

That's not quite right...look at the second term in the denominator again. :D

Once you fix this, then do the suggested division.

Second term should be \(\displaystyle n^{\frac{1}{3}}\) ??

 

[theakdad]

Second term should be \(\displaystyle n^{\frac{1}{3}}\) ??

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

 

[MarkFL]

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 \(\displaystyle \frac{n^{\frac{3}{2}}+1}{n+2}\) ??

Show me what you did to get that. It is wrong, but if I see what you did, I can address where the errors are.

 

[theakdad]

Show me what you did to get that. It is wrong, but if I see what you did, I can address where the errors are.

I have done addition for n in numerator and denominator.

 

[MarkFL]

I have done addition for n in numerator and denominator.

You cannot add because you do not have like terms.

 

[theakdad]

You cannot add because you do not have like terms.

I see,so i think then is so:

\(\displaystyle \frac{2+\frac{3}{4}+1}{1+\frac{1}{2}+2}\)

Am i right? If yes,then the solution is \(\displaystyle \frac{3\frac{3}{4}}{3\frac{1}{2}}\)

If this is ok then L= \(\displaystyle \infty\) because numerator is higher then denominator.

 

[MarkFL]

I see,so i think then is so:

\(\displaystyle \frac{2+\frac{3}{4}+1}{1+\frac{1}{2}+2}\)

Am i right? If yes,then the solution is \(\displaystyle \frac{3\frac{3}{4}}{3\frac{1}{2}}\)

If this is ok then L= \(\displaystyle \infty\) because numerator is higher then denominator.

How are you getting constants? Please let me see what you are doing. :D

 

[theakdad]

How are you getting constants? Please let me see what you are doing. :D

Those terms go all to infinity. I think so...

n² is infinity,all terms where n has exponent are infinity,so infinity + 1 or 2 is always infinity...

I suppose i was wrong when adding them together...
What do you say?

 

[MarkFL]

Those terms go all to infinity. I think so...

n² is infinity,all terms where n has exponent are infinity,so infinity + 1 or 2 is always infinity...

I suppose i was wrong when adding them together...
What do you say?

What does your limit look like after performing the divisions?

You need to write the expression in full; I can't really follow a shorthand of only exponents being written, if that is what you were doing. One of the keys to doing mathematics successfully, or at least to convey these ideas to others, is clearly written steps indicating the operations that have been performed. Your professor will most likely expect this on a test.

 

[theakdad]

What does your limit look like after performing the divisions?

You need to write the expression in full; I can't really follow a shorthand of only exponents being written, if that is what you were doing. One of the keys to doing mathematics successfully, or at least to convey these ideas to others, is clearly written steps indicating the operations that have been performed. Your professor will most likely expect this on a test.

Can u show me how should i write it?

 

[MarkFL]

Can u show me how should i write it?

You need to be able to do this yourself. If I show you how you should write it, then I have done the work, not you, and you gain relatively little from simply seeing it done compared to being able to do it yourself.

As the next step, rewrite the limit correctly with rational exponents, and I will guide you from there. :D

 

[theakdad]

You need to be able to do this yourself. If I show you how you should write it, then I have done the work, not you, and you gain relatively little from simply seeing it done compared to being able to do it yourself.

As the next step, rewrite the limit correctly with rational exponents, and I will guide you from there. :D

\(\displaystyle \frac{n^{\frac{4}{3}}+n^{\frac{1}{2}}+1}{n^{\frac{4}{6}}+n^{\frac{1}{3}}+2} \)

I thinh i have stuck here...

 

[MarkFL]

\(\displaystyle \frac{n^{\frac{4}{3}}+n^{\frac{1}{2}}+1}{n^{\frac{4}{6}}+n^{\frac{1}{3}}+2} \)

I thinh i have stuck here...

Good! Let's keep the limit notation too:

\(\displaystyle \lim_{n\to\infty}\frac{n^{\frac{4}{3}}+n^{\frac{1}{2}}+1}{n^{\frac{4}{6}}+n^{\frac{1}{3}}+2} \)

In the first term in ther denominator, I would reduce the rational exponent:

\(\displaystyle \lim_{n\to\infty}\frac{n^{\frac{4}{3}}+n^{\frac{1}{2}}+1}{n^{\frac{2}{3}}+n^{\frac{1}{3}}+2} \)

Now, divide each term in both the numerator and denominator by \(\displaystyle n^{\frac{2}{3}}\). This is the same as multiplying the entire expression of which we are taking the limit by one, so we are only changing its form, not its value.

Let's look at the first term in the numerator so divided:

\(\displaystyle \frac{n^{\frac{4}{3}}}{n^{\frac{2}{3}}}\)

To what does this simplify?