# Limit question

1. Jan 28, 2009

### transgalactic

2. Jan 28, 2009

### Gib Z

Before I start, I assume by "to take the lower integer part" you mean to use the floor function? ( http://en.wikipedia.org/wiki/Floor_function )

Well, it is true that 0^0 is an indeterminate form as the 2 variable limit

$$\lim_{(x,y) \to (0,0)} x^y$$ does not exist, but in this case we have the single variable limit:

$$\lim_{x\to 0} \lfloor (\sin x)^x \rfloor$$.

So lets ignore the floor function for a bit, and note that since $$\lim_{x\to 0} \frac{\sin x}{x} = 1$$, we can deduce that

$$\lim_{x\to 0} (\sin x)^x = \lim_{x\to 0} x^x$$.

Now, it is true that by direct substitution we get 0^0, an indeterminate form. But for this limit, rewrite $$x^x = \exp(x \log_e x)$$. You should know already that x "trumps" log x, ie has a greater influence in the limit, but if you don't we can see from its Taylor expansion that the function under exponentiation goes to 0. And so the limit is equal to 1.

But your limit is a bit more complicated with a floor function in it. Hopefully you can finish it off.