# How do you find the limit lim (2x+4)/(5-3x) as x->oo?

Jan 4, 2017

$= - \frac{2}{3}$

#### Explanation:

${\lim}_{x \to \infty} \frac{2 x + 4}{5 - 3 x}$

Divide numerator and deminitator by $x$ :)

$= {\lim}_{x \to \infty} \frac{2 + \frac{4}{x}}{\frac{5}{x} - 3}$

There are then various rules that allow you to do this, but it's also pretty intuitive:
$= \frac{2 + {\lim}_{x \to \infty} \frac{4}{x}}{{\lim}_{x \to \infty} \frac{5}{x} - 3}$

$= - \frac{2}{3}$

Jan 4, 2017

#### Explanation:

Given: ${\lim}_{x \rightarrow \infty} \frac{2 x + 4}{5 - 3 x}$

Because the expression evaluated at the limit is the indeterminate form, $- \frac{\infty}{\infty}$, then one should use L'Hôpital's rule .

Compute the derivative of the numerator:

$\frac{d \left(2 x + 4\right)}{\mathrm{dx}} = 2$

Compute the derivative of the denominator:

$\frac{d \left(5 - 3 x\right)}{\mathrm{dx}} = - 3$

Make a new expression:

${\lim}_{x \rightarrow \infty} \frac{2}{-} 3$

The limit of the above expression is easy to evaluate:

${\lim}_{x \rightarrow \infty} \frac{2}{-} 3 = - \frac{2}{3}$

The rule says that the limit of the original expression is the same:

${\lim}_{x \rightarrow \infty} \frac{2 x + 4}{5 - 3 x} = - \frac{2}{3}$