How do you find lim (3x^2+4)/(x^2-10x+25) as x->5?

Feb 27, 2017

See below.

Explanation:

First, plug in $x = 5$ to see if it is a problem:

....

$= \frac{3 {\left(5\right)}^{2} + 4}{{\left(5\right)}^{2} - 10 \left(5\right) + 25} = \text{ndef}$, because the denominator is zero :(

We can then see that:

${\lim}_{x \to 5} \frac{3 {x}^{2} + 4}{{x}^{2} - 10 x + 25}$

$= {\lim}_{x \to 5} \frac{3 {x}^{2} + 4}{{\left(x - 5\right)}^{2}}$

Clearly this hits $+ \infty$ as $x \to 5$ !! (NB: It's not a 2-sided limit, which is often an issue.)

As another approach, in these kinda problems, it can be useful to re-state the Origin, so we say that $z = x - 5$, or $x = z + 5$.

The problem becomes this:

${\lim}_{z \to 0} \frac{3 {\left(z + 5\right)}^{2} + 4}{{\left(z + 5\right)}^{2} - 10 \left(z + 5\right) + 25}$

$= {\lim}_{z \to 0} \frac{3 {z}^{2} + 30 z + 79}{{z}^{2}}$

$= {\lim}_{z \to 0} 3 + \frac{30}{z} + \frac{79}{z} ^ 2$

Same conclusion :)