How can I solve this limit $\lim_{x\to\infty} \frac{3x}{x + 4} = 3\,$?

Question

I have now $\forall \epsilon > 0\;\exists \delta > 0$, such that $\left|\frac{1}{x + 4}\right| < \frac{\epsilon}{12}$ if $x > \delta$. Now I don't know what I have to do. Can anyone help me?

Answer 1

Notice that $\delta > 0$ is not important in case $x\to \infty$.

Let $\epsilon>0$ be given. Define $\delta = \frac{12}{\min(\epsilon, 2)} - 4 \ge 2$. Then for every $x > \delta$ we have $$ \frac{3x}{x+4} - 3 = \frac{12}{x+4} < \frac{12}{\delta + 4} \le \min(\epsilon, 2) \le \epsilon. $$

Answer 2

Rewrite

$$ \frac{3x}{x+4} = \frac{3(x +4) }{x+4} -\frac{12}{x+4} = 3 - \frac{12}{x+4} $$ and use the fact you already have found.

Answer 3

$$\lim_{x\rightarrow\infty}\frac{3x}{x+4}=\lim_{x\rightarrow\infty}\frac{3x}{x(1+\frac{4}{x})}=\lim_{x\rightarrow\infty}\frac{3}{1+\frac{4}{x}}=3$$

Answer 4

I am trying to follow your thought and propose the following answer. First choose $\delta_1 = 1$ ,then if $x > 1=\delta_1 \to x+4 > x > 1 > 0 \to 0<\dfrac{1}{x+4} < \dfrac{1}{x}$. Next you want $\dfrac{1}{x} < \dfrac{\epsilon}{12}$, and this corresponds to choosing $\delta_2 = \dfrac{12}{\epsilon} > 0$. Now if you take $\delta = \text{max}(\delta_1,\delta_2) > 0$, then if $x > \delta \to \left|\dfrac{12}{x+4}\right| < \epsilon$, and this proves your claim.