limit problems! need help?

Question

I have several questions regarding to limits:
1. what is the limit of ln x as x approaches infinity?
2. how do you find the limit of (u^2+1)^1/2,as u approaches infinity?
3. can anyone tell me how to do the limit of cos (1/z) as z
approaches 0+ using the L' Hopital rules?

I need your guys help, this is not my homework but I just want to practice some limit problems! thanks

Answer 1

1) infinity
2) u^2 + 1 -> oo when u -> oo, therefore so does (u^2 + 1)^(1/2)
3) when z -0 , 1/z goes to + oo or -oo, depending on if you approach 0 from the right or from the left. Therefore, cos(1/z) has no limit at infinity. L'Hopital rule doesnt apply here.

Answer 2

1. As x goes to infinity, so does e^x, so as ln(e^x), which is x, goes to infinity, so does ln x.

Not good enough? All right, note that as e^x is increasing, also is ln x. Also notice that for any y, y=ln x has a solution in x, viz. e^y. Thus, y increases without bound, and the limit by definition is infinity.

2. As u goes to infinity, so do u^2, u^2 + 1, and sqrt(u^2 + 1), for reasons that may be familiar.

3. No one can tell you that. As z->0+, 1/z -> +inf., and cos (1/z) doesn't approach anything. It just bounces back and forth between -1 and +1.

HTH!

Answer 3

1. lim x→∞ lnx = ∞

2. lim u→∞ (u² + 1)^½
= lim u→∞ (u²)^½ (as for large u, u² + 1 → u²)
= lim u→∞ u
= ∞

3.limit z→0+ cos (1/z)
This is undefined as -1 ≤ cos u ≤ 1 for all u and 1/z →∞ as z→0+. So cos (1/z) oscillates

Answer 4

1. infinity
2. infinity again
3.you cannot use L'Hopital's rule in this case.
but the limit does not exist at all,
since
z->0 implies that 1/z->infinity and so
cos(1/z) just oscilates and does not converge to anything at all