# How do you evaluate lim_(x->3)(2x^2-5x-3)/(x-3)?

Nov 8, 2017

${\lim}_{x \rightarrow 3} \frac{2 {x}^{2} - 5 x - 3}{x - 3} = {\lim}_{x \rightarrow 3} \frac{\left(x - 3\right) \left(2 x + 1\right)}{x - 3} = {\lim}_{x \rightarrow 3} \left(2 x + 1\right) = 2 \left(3\right) + 1 = 7$

#### Explanation:

The numerator is $0$ when $x = 3$ ($3$ is a zero of the numerator).

Therefore $x - 3$ is a factor of the numerator. (So the quotient can be simplified for $x \ne 3$)

$2 {x}^{2} - 5 x - 3 = \left(x - 3\right) \left(\text{something}\right)$

By trial and error (or no error, just trial):

$2 {x}^{2} - 5 x - 3 = \left(x - 3\right) \left(2 x + 1\right)$

So $\frac{2 {x}^{2} - 5 x - 3}{x - 3} = \frac{\left(x - 3\right) \left(2 x + 1\right)}{x - 3} = 2 x + 1$ for $x \ne 3$.

The limit as $x \rightarrow 3$ doesn't care what happens right at $3$, just for $x$ close to $3$, so we can find the limit.

${\lim}_{x \rightarrow 3} \frac{2 {x}^{2} - 5 x - 3}{x - 3} = {\lim}_{x \rightarrow 3} \frac{\left(x - 3\right) \left(2 x + 1\right)}{x - 3} = {\lim}_{x \rightarrow 3} \left(2 x + 1\right) = 2 \left(3\right) + 1 = 7$

Nov 8, 2017

The function approaches 7

#### Explanation:

To find the limit of a function, we use L'Hopital's rule, where: ${\lim}_{x \to a} f \frac{x}{g} \left(x\right) = \frac{f ' \left(x\right)}{g ' \left(x\right)} , a \ne \setminus \infty$ then substitute $a$ for $x$.

So, for our example: ${\lim}_{x \to 3} \frac{2 {x}^{2} - 5 x - 3}{x - 3}$, to find the limit we differentiate both equations to get:
${\lim}_{x \to 3} \frac{4 x - 5}{1}$

Now we replace $x$ with 3:
${\lim}_{x \to 3} \frac{4 \left(3\right) - 5}{1} = 7$