Calculate limx→∞(sin√x+1−sin√x).
Let:
L=limx→∞(sin(√x+1)−sin(√x))
Application of the sum-to-product identity:
sin(α)−sin(β)=2sin(α−β2)cos(α+β2)
And the limit property:
limx→c(k⋅f(x))=k⋅limx→c(f(x))
Allows us to write:
L=2limx→∞(sin(√x+1−√x2)cos(√x+1+√x2))
Rationalization of the numerator of the sine function gives us:
L=2limx→∞(sin(12(√x+1+√x))cos(√x+1+√x2))
Now, using the property of limits:
limx→c(f(x)⋅g(x))=(limx→c(f(x)))(limx→c(g(x)))
We obtain:
L=2limx→∞(sin(12(√x+1+√x)))limx→∞(cos(√x+1+√x2))
The first limit goes to zero, and the second limit is bounded, hence:
L=0
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