1. May 3, 2009

### C.E

1. Find the following limits, if they exist, justify your answers fully. ([x] denotes the integer part of x).

a. $$\[lim_{x \to 1+} \frac{[x]}{x^2}$$

b.$$\[lim_{x \to 1-} \frac{[x]-x}{[x^2]-x^2}$$

c.$$\[lim_{x \to \infty}$$(1/x)^(1/x)

d.$$\[lim_{x \to 0} \frac{exp(2x)-1}{ln(1+x)}$$

e.$$\[lim_{x \to 0} \frac{x^2sin(1/x)}{sin(x)}$$

3. The attempt at a solution.

I am really stuck on the first two and don't know how to start, I am confused about how to deal with the integer parts in the limits. Could someone please offer some guidance?

For d, I used L Hopitals rule and got a limit of 2, is this correct? For e, I tried to use L Hopitals rule and think I managed to show that the limit was the same as the limit of 2sin(1/x) -cos(1/x)/x^3 and hence that it does not exist. Have I done these right?

2. May 3, 2009

### xaos

for the first two, try the epsilon delta definition of a limit. the floor function and the 1/x^2 function will both be bounded below by zero (infact [x]=1 for all positive delta!), so that you can multiply them without error.

for the last one, try to rewrite this into the product of two well known limits.

3. May 3, 2009

### C.E

By the way the part a should have been the limit of [x]/[x^2], in which case could I just say the limit is one and to prove it let delta =0.1 givng f(x)-1=0. I can also see how to do the last part now but am really stuck on b. Could somebody please give me a hint about what the limit is and which delta value I should use to prove it?

4. May 3, 2009

### HallsofIvy

Staff Emeritus
For (b), since the limit is taken as x approaches 1 from below, you can assume that x is just slightly less than 1. In that case, both [x] and $[x^2]$ are 0. The limit is exactly the same as $\lim_{x\rightarrow 1^-}-x/-x^2= \lim_{x\rightarrow 1^-} 1/x$

5. May 3, 2009

### C.E

Does anybody know how to do part c?

6. May 3, 2009

### Staff: Mentor

For #3, let y = (1/x)^(1/x), then take logs of both sides, so that
ln y = 1/x ln(1/x)
Now take the limit as x --> infinity, and L'Hopital's Rule is your friend.
After you have the limit, what you have is the limit of the ln of what you want, so the limit of the original problem is e raised to that power.