## anonymous one year ago if lim(x-->0)f(x)=-1find the lim(x-->0) (f(x)+1)sin(pi/x^2)

1. jtvatsim

Could we use the squeeze theorem on this one?

2. jtvatsim

Consider that $(f(x) + 1) \cdot -1 \leq (f(x) + 1) \cdot \sin(\pi/x^2) \leq (f(x) + 1) \cdot 1$ hence, $\lim_{x\rightarrow0}(f(x) + 1) \cdot -1 \leq \lim_{x\rightarrow0}(f(x) + 1) \cdot \sin(\pi/x^2) \leq \lim_{x\rightarrow0}(f(x) + 1) \cdot 1$ that is, $0\leq \lim_{x\rightarrow0}(f(x)+1) \cdot \sin(\pi/x^2) \leq 0$ and our desired limit is simply 0.

3. jtvatsim

The sin(pi/x^2) feels uncomfortable since we "know" that the inside is heading toward something undefined (or infinite), but the sin function itself is always stuck between -1 and 1 no matter what the inside is.