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Let
f(x)=[(sin[x])/(([x]+1)²),, x>0,; [6pt]
(cos ((π)/(2)[x]))/([x]),, x<0,]
where [x] denotes the greatest integer not exceeding x. For f to be continuous at x=0, the value of k must be
- 0
- 1
- -1
- indeterminate
Correct answer: 0
Solution
For x->0+, [x]=0 so sin([x])/([x]+1)^2 = sin0/1 = 0. For x->0-, [x]=-1 so cos((pi/2)[x])/[x] = cos(-pi/2)/(-1) = 0/(-1) = 0. Both one-sided limits equal 0, so f(0) = k = 0 for continuity.
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