Answer: [tex]\frac{-m^2+n^2}{2}[/tex]
If the font is too small, it says (-m^2+n^2)/2
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Explanation:
If you plug x = 0 into the original expression, then you'll get the indeterminant form 0/0 after simplifying.
Recall that cos(0) = 1
So we use L'Hospitals rule to derive the numerator and denominator separately (don't use the quotient rule)
[tex]\displaystyle \lim_{x\to0} \frac{\cos(mx)-\cos(nx)}{x^2}\\\\\\\displaystyle \lim_{x\to0} \frac{-m\sin(mx)+n\sin(nx)}{2x}\\\\\\[/tex]
Plug x = 0 into this new expression. You should get 0/0 again here as well, which means you'll have to do another round of L'Hospital
[tex]\displaystyle \lim_{x\to0} \frac{-m^2\cos(mx)+n^2\cos(nx)}{2}\\\\\\[/tex]
Now plugging in x = 0 leads to something that isn't indeterminant
Evaluating the limit gets us
[tex]\displaystyle \lim_{x\to0} \frac{-m^2\cos(mx)+n^2\cos(nx)}{2}\\\\\\=\frac{-m^2\cos(m*0)+n^2\cos(n*0)}{2}\\\\\\= \frac{-m^2+n^2}{2}\\\\\\[/tex]
