# Question #06261

Oct 14, 2017

It is because ${\lim}_{x \to \infty} \left(\frac{1}{x}\right) = 0$

#### Explanation:

I assume the question is written as:
${\lim}_{x \to \infty} \left(\frac{6}{x}\right) - 3$

We can first factor out the $6$ like so:
${\lim}_{x \to \infty} \left(\frac{6}{x}\right) - 3$
$= {\lim}_{x \to \infty} \left(6 \cdot \frac{1}{x}\right) - 3$
$= {\lim}_{x \to \infty} \left(6\right) \cdot {\lim}_{x \to \infty} \left(\frac{1}{x}\right) - 3$
$= 6 \cdot {\lim}_{x \to \infty} \left(\frac{1}{x}\right) - 3$

Then, we just solve for:

${\lim}_{x \to \infty} \left(\frac{1}{x}\right)$

Recall that the limit of $\frac{1}{x}$ as $x$ approaches $\infty$ is $0$.
This is because, for bigger and bigger values of $x$, $\frac{1}{x}$ becomes smaller and smaller.

graph{1/x [-4.25, 15.75, -0.56, 9.44]}

You can see on the above graph, that as you go further along the x-axis, the closer the graph gets to $0$.

So, the limit of this graph as $x$ approaches $\infty$ is $0$.

Now, we know that:
${\lim}_{x \to \infty} \left(\frac{1}{x}\right) = 0$

So:

$6 \cdot {\lim}_{x \to \infty} \left(\frac{1}{x}\right) - 3$

$= 6 \cdot 0 - 3$

$= - 3$