limit goes to infinity [sqrt(2) + 3x^-3/2 + 2sqrt(x) / x]

[Jade]

I need to evaluate the limit only

[sqrt(2) + 3x^-3/2 + 2sqrt(x) / x]

I have come up with 3

 

[stapel]

Your subject line says that the limit goes to infinity, but your post says that the limit is 3. Which is it? Or did you mean that the limit is to be taken as x goes to infinity, and the limit is 3?

Why do you think your answer is incorrect? What steps did you use to get that answer?

Thank you.

Eliz.

 

[galactus]

Of course, no parentheses. Your post is ambiguous. Do you mean:

\(\displaystyle \L\\\lim_{x\to\infty}\frac{\sqrt{2}+3x^{\frac{-3}{2}}+2\sqrt{x}}{x}\)

\(\displaystyle \L\\\lim_{x\to\infty}\frac{\sqrt{2}}{x}+\lim_{x\to\infty}\frac{3}{x^{\frac{5}{2}}}+\lim_{x\to\infty}\frac{2}{\sqrt{x}}\)

The limit is not 3. Try again. Can you see it?.

 

[Jade]

sorry

limit x goes to infinity

You almost decifered my code

the entire equation is not divided by x, just the last part 2 sqrt(x) is over x

Let me try to write it again [sqrt(2) + 3x^-3/2 + (2 sqrt(x)/x)]

Does this make sense??

 

[stapel]

limit x goes to infinity

the entire equation is not divided by x, just the last part 2 sqrt(x) is over x

Is the exercise as follows?

. . . . .\(\displaystyle \L \begin{array}{c}\mbox{limit}\\x\rightarrow \infty \end{array}\,\, \sqrt{2}\, +\, 3x^{-\frac{3}{2}}\, +\, \frac{2\sqrt{x}}{x}\)

Thank you.

Eliz.

 

[Jade]

You got it!

Sorry for the confusion

 

[galactus]

You have the limit:

\(\displaystyle \L\\\lim_{x\to\infty}\sqrt{2}+\lim_{x\to\infty}\frac{3}{x^{\frac{3}{2}}}+\lim_{x\to\infty}\frac{2}{\sqrt{x}}\)


See the limit?. It's in front of you.

 

[Jade]

I hope I am not wrong

I am afraid to let out my answer in fear that it is incorrect

One

 

[Jade]

My thoughts

I know that
1/x=0
1/x^2=0
1/sqrt(x)=0

I am thinking that the first part is 1 sqrt(2)

This is where I am coming up with the one

 

[stapel]

Re: My thoughts

I know that 1/x=0

What do you mean by this? Since 1 does not equal zero, then this fraction, 1/x, can never equal zero. Do you perhaps mean "the limit of 1/x, as x tends toward infinity, is zero"?

1/x^2=0
1/sqrt(x)=0

Same question.

I am thinking that the first part is 1 sqrt(2)

Why do you need an extra "1"?

Yes, the limit of a constant term, as the variable changes, is the value of the constant term. Now what have you done with the other two terms, 3/(xsqrt[x]) and 2/sqrt[x]?

Thank you.

Eliz.

 

[Jade]

I did mean that the limit of 1/x, as x tends toward infinity, is zero

I am trying to go through my rules to come to my evaluation. This is all new to me.

I am guessing I am way off.

 

[galactus]

Re: I hope I am not wrong

I am afraid to let out my answer in fear that it is incorrect

One

Don't be afraid. "Fear is the mind-killer"(Paul Atreides).

No. This limit is rather intuitive by comparison. If you have 1/x and x gets increasingly large, what does it approach?. Think about it.

What is 1/1000, 1/1000000, 1/1000000000000000000000000?

It gets closer to 0, doesn't it?. I broke your limit up into 3 limits. What does the above tell you about the 2 rightmost limits?. What is left?.
Which limit has no x?.

 

[Jade]

The two most right limits

The two most right limits are equal to zero

The only one without the x is sqrt(2)

 

[galactus]

There ya' go!.

See how that works?. Mind you, not all are that intuitive, but do you see what is going on?.

 

[Jade]

I think I am really getting it

thanks