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Siobhan
Siobhan

Math chevron_right United states

CALCULUS LIMITS :(

Date Posted: 2 months ago
Views: 25
J
J
2 months ago
Use the squeeze theorem.

-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+