Consider the function $$f(x) = [x + 1] - \sin\left(\frac{\pi}{2}[x]\right)$$ for $$x \epsilon R$$, where [x] denotes the greatest integer less than or equal to x. Let $$l_1 = \lim_{x \rightarrow 0^-}f(x)$$ and $$l_2 = \lim_{x \rightarrow 0^+}f(x)$$ It follows that

$$l_1 = \lim_{x \rightarrow 0^-}f(x)$$

When $$x\longrightarrow\ 0^-$$ then $$\left[x\right]$$ = -1

f(x) can be written as $$[x]+1-\sin\ \left(\ \frac{\pi\ \ }{2}\left[x\right]\right)$$.

Putting [x] = -1 we get $$l_1$$ = $$-1+1-\sin\ \left(\ \frac{\ \pi\ }{2}\times\ \left(-1\right)\ \right)$$ = $$-\sin\ \left(\ \frac{\ \pi\ }{2}\times\ \left(-1\right)\ \right)$$ = 1

for $$l_2$$, $$x\longrightarrow\ 0^+$$ thus [x] = 0

putting it in f(x). $$l_2$$ = 0+1-0 = 1

Hence $$l_1 =l_2=1$$