Ask Singapore Homework?
Upload a photo of a Singapore homework and someone will email you the solution for free.
Question

No Answers Yet
Help Siobhan! Anyone can contribute an answer, even nontutors.
CALCULUS LIMITS :(
1 ≤ sin(π/x) ≤ 1
So lim (√x e^(1)) ≤ lim (√x e^(sin(π/x)) ≤ lim (√x e¹)
Now when x = 0, both (√x e^(1)) and (√x e¹) = 0
So lim for both is 0.
Therefore by the squeeze theorem,
lim (√x e^(sin(π/x)) = 0 when x→ 0+