Question

Can someone please help me with this by doing L hospitals rule?

Answer 1

Answer:
`(-m^2+n^2)/(2)`

If the font is too small, it says (-m^2+n^2)/2

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Step-by-step 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)

`\displaystyle \lim_(x\to0) (\cos(mx)-\cos(nx))/(x^2)\\\\\\\displaystyle \lim_(x\to0) (-m\sin(mx)+n\sin(nx))/(2x)\\\\\\`

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

`\displaystyle \lim_(x\to0) (-m^2\cos(mx)+n^2\cos(nx))/(2)\\\\\\`

Now plugging in x = 0 leads to something that isn't indeterminant

Evaluating the limit gets us

`\displaystyle \lim_(x\to0) (-m^2\cos(mx)+n^2\cos(nx))/(2)\\\\\\=(-m^2\cos(m*0)+n^2\cos(n*0))/(2)\\\\\\= (-m^2+n^2)/(2)\\\\\\`