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CALCULUS LIMITS :(
Date Posted: 5 years ago
Views: 465
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+
-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+
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