Question

lim x tans to 0 (sinax /e^bx - 1)




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Answer 1

Answer:

Step-by-step explanation:

Here is another solution. Convert the denominator to product using trigonometric formulas. Now divide both numerator and denominator by ax-bx. But we should use the fact that e ^ax/ ax ---> 1 where the latter follows from the Taylor series expansion for the function f(x)=eax. So L=a−ba−b=1.

eax−ebx=(a−b)x+o(x2) , and sin(ax)−sin(bx)=2sin((a−b)x2)cos((a+b)x2). So:

eax−ebxsin(ax)−sin(bx)=(a−b)x+o(x2)(a−b)x+o(x3)=a−b+o(x)a−b+o(x2), from this the limit is seen to be a−ba−b=1