Calculate the limit of limx→11−x2sin⁡πx only with the fundamental limits

Let t=x−1. Then we have 1−x2sin⁡(πx)=−t(t+2)sin⁡(π(t+1))=t(t+2)sin⁡(πt) Now, set πt=y to find limx→11−x2sin⁡(πx)=limy→0(ππ(yπ+2)ysin⁡y)

Resolved: "Calculate the limit of limx→11−x2sin⁡πx only with the..."

Carol Valentine

2022-01-15

Calculate the limit of

lim_{x \to 1} \frac{1 - x^{2}}{\sin π x}

only with the fundamental limits

amarantha41

Beginner2022-01-16Added 38 answers

Let t=x−1. Then we have

\frac{1 - x^{2}}{\sin (π x)} = \frac{- t (t + 2)}{\sin (π (t + 1))} = \frac{t (t + 2)}{\sin (π t)}

Now, set

π t = y

to find

lim_{x \to 1} \frac{1 - x^{2}}{\sin (π x)} = lim_{y \to 0} (\frac{π}{π} (\frac{y}{π} + 2) \frac{y}{\sin y})

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