Formal limit definition when x tends to infinity

Question

Using the limit definition, show that:

$\lim\limits_{x\to\infty} \frac{x+7}{3x^2+2}=0$

I get blocked when I use the equation:

Definition $f(x)$ tends to $L$ as $x$ tends to $ \infty$ if and only if $$ \forall \epsilon >0, \exists \delta>0 \ \mbox{ such that} \ \forall x \ \mbox{where} \ x>\delta, \\ |f(x)-L|<\epsilon$$

if I apply the equation would look like this:

$\frac{x+7}{3x^2+2}-0 < ε$

but I do not know how to continue

Answer 1

Here's one way to do it. Let's simplify the fraction a bit first with some inequalities. First, the denominator:

$$f(x) = \frac{x+7}{3x^2+2} < \frac{x+7}{3x^2}.$$

Now, for the numerator, notice that $x+7 < x+7x$ provided that $x>1$. So, continuing from above, we have

$$f(x) < \frac{x+7}{3x^2} < \frac{x+7x}{3x^2} = \frac{8x}{3x^2} = \frac8{3x}.$$

We want to guarantee that $f(x) < \epsilon$, so it is enough to guarantee that $\frac8{3x} < \epsilon$. Solving this last inequality for $x$, we have

$$x > \frac8{3\epsilon}.$$

Therefore we should choose any $\delta$ such that $$\delta > \max\left\{\frac8{3\epsilon},1\right\}.$$

(Recall that we need to make sure that $x>1$, which is why we're taking the maximum of $\frac8{3\epsilon}$ and $1$.)

Answer 2

You solve the inequation for $x$, thus obtaining $x\in (-\infty ,\alpha(\varepsilon))\cup (\beta(\varepsilon),\infty)$, and $\beta(\varepsilon)$ will be a good candidate for $\delta$.