Evaluate limx→πsin⁡(mx)sin⁡(nx) where m,n∈N⋅. At first I thought I could just use the remarkable limit limx→0sin⁡(x)x=1 and the answer could just be mn but this is not the answer.... I mean it's a part of it but I don't understand why.

Let y=x−π. Thensin⁡(mx)=sin⁡(m(y+π))=(−1)msin⁡(my)∼(−1)mmythus, the limit is(−1)m−nmn

HINTLet x=y+π thenlimx→πsin⁡(mx)sin⁡(nx)=limy→0sin⁡(mπ+my)sin⁡(nπ+ny)then consider the cases with m,n odd or even.

Let y=x−π. Thensin⁡(mx)=sin⁡(m(y+π))=(−1)msin⁡(my)∼(−1)mmythus, the limit is(−1)m−nmn

HINTLet x=y+π thenlimx→πsin⁡(mx)sin⁡(nx)=limy→0sin⁡(mπ+my)sin⁡(nπ+ny)then consider the cases with m,n odd or even.

Best solutions to - "Evaluate limx→πsin⁡(mx)sin⁡(nx) where m,n∈N⋅. At first I thought..."

talpajocotefnf3

2022-04-05

Evaluate

lim_{x \to π} \frac{\sin (m x)}{\sin (n x)}

where

m, n \in N \cdot

. At first I thought I could just use the remarkable limit

lim_{x \to 0} \frac{\sin (x)}{x} = 1

and the answer could just be

\frac{m}{n}

but this is not the answer.... I mean it's a part of it but I don't understand why.

Abdullah Avery

Beginner2022-04-06Added 19 answers

Let

y = x - π

. Then

\sin (m x) = \sin (m (y + π))

= {(- 1)}^{m} \sin (m y) \sim {(- 1)}^{m} m y

thus, the limit is

{(- 1)}^{m - n} \frac{m}{n}

Alannah Campos

Beginner2022-04-07Added 10 answers

HINT
Let

x = y + π

then

lim_{x \to π} \frac{\sin (m x)}{\sin (n x)} = lim_{y \to 0} \frac{\sin (m π + m y)}{\sin (n π + n y)}

then consider the cases with m,n odd or even.

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