Welcome to our community

Be a part of something great, join today!

[SOLVED] Another two limits at infinity

wishmaster

Active member
Oct 11, 2013
211
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
Jan 26, 2012
995
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

Administrator
Staff member
Feb 24, 2012
13,775
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
Oct 11, 2013
211
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

Administrator
Staff member
Feb 24, 2012
13,775
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
Oct 11, 2013
211
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

Administrator
Staff member
Feb 24, 2012
13,775
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
Oct 11, 2013
211
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
Jan 26, 2012
995
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 $\displaystyle \lim_{n\to\infty}\frac{8}{5}\left( \frac{5}{3}\right)^n$.
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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.
 

wishmaster

Active member
Oct 11, 2013
211
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 ;)
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
May I ask the OP what course is he taking that discusses these limits ?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
...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

Administrator
Staff member
Feb 24, 2012
13,775

wishmaster

Active member
Oct 11, 2013
211
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

Administrator
Staff member
Feb 24, 2012
13,775
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
Oct 11, 2013
211
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
Jan 26, 2012
995
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

Administrator
Staff member
Feb 24, 2012
13,775
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
Oct 11, 2013
211
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?
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
\(\displaystyle \frac{n^{\frac{4}{3}}+n^{\frac{1}{2}}+1}{n^{\frac{4}{6}}+n^{\frac{2}{3}}+2} \)
That's not quite right...look at the second term in the denominator again. :D

Once you fix this, then do the suggested division.
 

wishmaster

Active member
Oct 11, 2013
211
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 \(\displaystyle n^{\frac{1}{3}}\) ??
 

wishmaster

Active member
Oct 11, 2013
211

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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

Administrator
Staff member
Feb 24, 2012
13,775
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.