Find limit of function without l'Hospital method.

Question

Find $\displaystyle \lim_{x\to 8}\left(\frac{8-x}{\sin \frac{\pi x}{8}}\right)$.

$t=x-8 => x=t+8$

$\displaystyle \lim_{t\to 0}\left(\frac{t}{\sin \frac{\pi(t+8) }{8}}\right)$ = $\displaystyle \lim_{t\to 0}\left(\frac{t}{-\sin \frac{t\pi }{8}}\right)$ = $\displaystyle \lim_{\frac{t\pi }{8}\to 0}\left(\frac{t}{-\sin \frac{t\pi }{8}}\right)$ = ...

Is this correct method?

Answer 1

yes,you should just conclude that $$\lim _{ t\to 0 } \left( \frac { t }{ -\sin \frac { t\pi }{ 8 } } \right) =\lim _{ t\to 0 } \left( \frac { \frac { t\pi }{ 8 } }{ -\sin \frac { t\pi }{ 8 } } \right) \frac { 8 }{ \pi } =-\frac { 8 }{ \pi } $$

Answer 2

Yes your approach works. Another approach: Set $f(x) = \sin (\pi x/8).$ Then the expression equals

$$-\frac{x-8}{f(x) - f(8)} = -\left (\frac{f(x) - f(8)}{x-8} \right )^{-1}.$$

As $x\to 8,$ the expression inside the parentheses tends to $ f'(8)$ by the definition of the derivative. The rest is a simple computation.