MHB Another two limits at infinity

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The discussion centers around evaluating two limits as n approaches infinity. The first limit involves the expression with exponential terms, where participants suggest dividing by the dominant term to simplify the evaluation. The second limit involves radical expressions, with advice given to convert them to rational exponent notation for easier manipulation. Participants emphasize the importance of clearly writing out each step in the evaluation process to avoid confusion and ensure understanding. Ultimately, the conversation highlights strategies for handling limits and the significance of proper notation in mathematical communication.
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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.
[math]\lim _{n \to \infty} \frac{5^{n^+1}-2*5^n+5^{n-1}}{3^{n+1}-3^n}[/math]

2.
[math]\lim _{n \to \infty} \frac{\sqrt[3]{n^4}+\sqrt{n}+1}{\sqrt[6]{n^4}+\sqrt[3]{n}+2}[/math]
 
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wishmaster said:
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.
[math]\lim _{n \to \infty} \frac{5^{n^+1}-2*5^n+5^{n-1}}{3^{n+1}-3^n}[/math]

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

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

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

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

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

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

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

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

to obtain:

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

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

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

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

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

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

Converting from logarithmic to exponential form, we have:

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

If $b=1$ then we have:

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

Converting from logarithmic to exponential form:

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

If $1<b$ then we have:

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

Converting from logarithmic to exponential form, we find:

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

What do you find?
 
MarkFL said:
1.) By factoring, you can get this limit to the form:

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

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

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

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

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

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

to obtain:

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

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

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

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

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

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

Converting from logarithmic to exponential form, we have:

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

If $b=1$ then we have:

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

Converting from logarithmic to exponential form:

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

If $1<b$ then we have:

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

Converting from logarithmic to exponential form, we find:

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

What do you find?

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

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

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

How would you respond to these?
 
MarkFL said:
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:

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

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

How would you respond to these?

They both go to infinity?
 
wishmaster said:
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?
 
MarkFL said:
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 $$\frac{2}{3}$$ is going to zero,while $$\frac{3}{2}$$ is going to infinity
 
wishmaster said:
Smaller.

So $$\frac{2}{3}$$ is going to zero,while $$\frac{3}{2}$$ 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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  • #10
wishmaster said:
Smaller.

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

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

If we write the expression as:

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

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:

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

And then observe that this ratio gets larger and larger as $n$ grows without bound, we then conclude that the limit goes to infinity.
 
  • #11
MarkFL said:
Yes...but see the post above by Chris L T521 for clarification. :D

If we write the expression as:

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

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:

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

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 ;)
 
  • #12
May I ask the OP what course is he taking that discusses these limits ?
 
  • #13
Chris L T521 said:
...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:

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

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

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

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

$$L=\frac{8}{5}\lim _{n\to\infty} \left(\frac{5}{3} \right)^n$$
 
  • #14
wishmaster said:
i understand that...question is how to solve limits ;)

Understanding this is key to evaluating the given limit.
 
  • #15
MarkFL said:
I get a slightly different result:

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

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

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

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

$$L=\frac{8}{5}\lim _{n\to\infty} \left(\frac{5}{3} \right)^n$$
So in numerator you have exposed $$5^{n-1}$$ and in denominator $$3^n$$?
 
  • #16
wishmaster said:
So in numerator you have exposed $$5^{n-1}$$ and in denominator $$3^n$$?

Yes, you want to expose (or factor out) the smallest power, because it is a common factor to all terms.
 
  • #17
MarkFL said:
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?
 
  • #18
MarkFL said:
I get a slightly different result:

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

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

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

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

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

Woops, I misread my own handwriting... (Headbang)

I got the same thing you did. XD
 
  • #19
wishmaster said:
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:

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

and then use the property of exponents:

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

when doing the suggested divisions. What do you get?
 
  • #20
MarkFL said:
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:

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

and then use the property of exponents:

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

when doing the suggested divisions. What do you get?

$$\frac{n^{\frac{4}{3}}+n^{\frac{1}{2}}+1}{n^{\frac{4}{6}}+n^{\frac{2}{3}}+2} $$

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

Am i right?
 
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  • #21
wishmaster said:
$$\frac{n^{\frac{4}{3}}+n^{\frac{1}{2}}+1}{n^{\frac{4}{6}}+n^{\frac{2}{3}}+2} $$

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

Once you fix this, then do the suggested division.
 
  • #22
MarkFL said:
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 $$n^{\frac{1}{3}}$$ ??
 
  • #23
wishmaster said:
Second term should be $$n^{\frac{1}{3}}$$ ??

Then is $$\frac{n^{\frac{3}{2}}+1}{n+2}$$ ??
 
  • #24
wishmaster said:
Second term should be $$n^{\frac{1}{3}}$$ ??

Correct:

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

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

$$n^{\frac{4}{6}}=n^{\frac{2}{3}}$$ ?
 
  • #25
wishmaster said:
Then is $$\frac{n^{\frac{3}{2}}+1}{n+2}$$ ??

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.
 
  • #26
MarkFL said:
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.
 
  • #27
wishmaster said:
I have done addition for n in numerator and denominator.

You cannot add because you do not have like terms.
 
  • #28
MarkFL said:
You cannot add because you do not have like terms.

I see,so i think then is so:

$$\frac{2+\frac{3}{4}+1}{1+\frac{1}{2}+2}$$

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

If this is ok then L= $$\infty$$ because numerator is higher then denominator.
 
  • #29
wishmaster said:
I see,so i think then is so:

$$\frac{2+\frac{3}{4}+1}{1+\frac{1}{2}+2}$$

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

If this is ok then L= $$\infty$$ because numerator is higher then denominator.

How are you getting constants? Please let me see what you are doing. :D
 
  • #30
MarkFL said:
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?
 
Last edited:

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