$$\lim_{x \to 0} \log_4 \frac{4-x}{x+1} =1$$
Prove the limit by $\epsilon-\delta$ proof.
The best I could do is narrowing it to $\frac{4-x}{4(x+1)}<4^\epsilon$ using logarithm properties, but if I try to isolate x I end up with $$x>-\frac{4(4^\epsilon-1)}{1+4^{\epsilon+1}}$$ which doesn't lead me anywhere.
Any help would be appreciated!
Using the definition we have
$$\vert \log_4\left(\dfrac{4-x}{x+1}\right) - 1\vert < \epsilon$$
that is
$$-\epsilon < \log_4\left(\dfrac{4-x}{x+1}\right) -1 < \epsilon$$
$$4^{-\epsilon+1} < \dfrac{4-x}{x+1} < 4^{\epsilon+1}$$
Now with some tedious little algebra:
$$4^{-\epsilon+1}(x+1) < 4-x < 4^{\epsilon+1}(x+1)$$
$$4^{-\epsilon+1} - 4 < -4^{-\epsilon+1}x - x - 4^{\epsilon+1}x < 4^{\epsilon+1} - 4$$
$$4^{-\epsilon+1} - 4 < x(-4^{-\epsilon+1} - 1 - 4^{\epsilon+1}) < 4^{\epsilon+1} - 4$$
Whence
$$\dfrac{4^{-\epsilon+1} - 4}{(-4^{-\epsilon+1} - 1 - 4^{\epsilon+1})} < x < \dfrac{4^{\epsilon+1} - 4}{(-4^{-\epsilon+1} - 1 - 4^{\epsilon+1})}$$
Now as $\epsilon \to 0$ we are creating a neighbourhood of $0$, hence the limit is verified.