Limit of function with square roots

[teng125]

for sqr root of (n + sqr root (n) ) - sqr root (n),is the answer = zero or infinity so converges or diverges??

 

[EnumaElish]

What would be the argument as to why it should be zero?

 

[teng125]

if i divide each constant n by n itself then limit n to infinity and i'll get 1 - 1 =0 right??

 

[EnumaElish]

You mean, you'll have

$$n\left(\sqrt{\frac1n+\sqrt{\frac1{n^3}}}-\sqrt{\frac1n}}\right)$$

and since 1/n³ ---> 0 much faster than 1/n, you'll end up with n[sqrt(1/n) - sqrt(1/n)]?

 

[teng125]

no,just sqr root of 1/n without the power of 3

 

[teng125]

if i divide every thing by n,i'll get sqr root [1 + (sqr root 1/n)] - sqr root (1)
what is the final answer??

 

[HallsofIvy]

If you divide each term of
$$\sqrt{n+ \sqrt{n}}-\sqrt{n}$$
by $\sqrt{n}$ you get
$$\sqrt{1+ \frac{1}{\sqrt{n}}}- 1$$
What is the limit of that as n goes to infinity?


By the way, why are you posting so many problems involving limits, derivatives, and integrals in the precalculus section?

 

[teng125]

is the answer = to zero??

 

[VietDao29]

I have a question. It's not clear to tell that what you are asking for. Are you asking for this:
$$\lim_{n \rightarrow \infty} \sqrt{n - \sqrt{n}} - \sqrt{n}$$
or this:
$$\lim_{n \rightarrow \infty} \frac{\sqrt{n - \sqrt{n}} - \sqrt{n}}{\sqrt{n}}$$?
If it's the latter, then you are correct!