MHB 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.
[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]

 

[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.
[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?

 

[MarkFL]

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?

 

[theakdad]

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...

 

[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:

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

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

How would you respond to these?

 

[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:

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

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

How would you respond to these?

They both go to infinity?

 

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

 

[Chris L T521]

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

 

[MarkFL]

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.

 

[theakdad]

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 ;)

 

[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:

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

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

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

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

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

 

[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:

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

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

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

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

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

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

 

[MarkFL]

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.

 

[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:

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

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

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

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

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

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

I got the same thing you did. XD

 

[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:

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

and then use the property of exponents:

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

when doing the suggested divisions. What do you get?

 

[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:

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

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?

 

[MarkFL]

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

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 $$n^{\frac{1}{3}}$$ ??

 

[theakdad]

Second term should be $$n^{\frac{1}{3}}$$ ??

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

 

[MarkFL]

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:

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

 

[MarkFL]

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.

 

[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:

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

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.

 

[MarkFL]

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

 

[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?