Question

Preciso de ajuda com os seguintes limites:
(x+3)(x-4)/(x+3)(x+1)
x tendendo a -3

senh(x)/2
x tendendo a 2

Answer 1

Temos:

$\lim_{x \to -3} \frac{(x + 3)(x - 4)}{(x + 3)(x + 1)} =\\\\\lim_{x \to -3} \frac{(x - 4)}{(x + 1)} =\\\\\frac{-3 - 4}{-3 + 1} =\\\\\frac{7}{2}$

$\lim_{x \to 2} \frac{senh(x)}{2} =\\\\\lim_{x \to 2} \frac{\frac{e^x - e^{-x}}{2}}{2} =\\\\\lim_{x \to 2} \frac{e^x - e^{-x}}{4} =\\\\\frac{e^2 - e^{-2}}{4} =\\\\\frac{e^2 - 1/e^2}{4} =\\\\\frac{\frac{e^4 - 1}{e^2}}{4} =\\\\\frac{e^4 - 1}{4e^2}$

Answer 2

a)

Como não ficou claro, vou assumir que (x+3)(x-4) sejam numeradores e (x+3)(x+1) sejam denominadores.

$\underset{x\to-3}{lim}~\frac{(x+3).(x-4)}{(x+3).(x+1)}\\\\\\\underset{x\to-3}{lim}~\frac{(x+3)\!\!\!\!\!\!\!\!\!\!\!\!\!{////}.(x-4)}{(x+3)\!\!\!\!\!\!\!\!\!\!\!\!\!{////}.(x+1)}\\\\\\\underset{x\to-3}{lim}~\frac{(x-4)}{(x+1)}~=~\frac{(-3-4)}{(-3+1)}~=~\frac{-7}{-2}~=~\boxed{3,5}$

b)

Utilizando a forma exponencial de sinh, temos:

$\underset{x\to2}{lim}~\frac{sinh(x)}{2}\\\\\\\underset{x\to2}{lim}~\frac{\frac{e^x-e^{-x}}{2}}{2}\\\\\\\underset{x\to2}{lim}~\frac{e^x-e^{-x}}{4}~=~\frac{e^2-e^{-2}}{4}~\approx~\boxed{1,81}$