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

Nov 2, 2016

${\lim}_{x \to - \infty} \frac{{x}^{2} + 3}{{x}^{2} + 4} = 1$

#### Explanation:

If we look at the graph of $y = \frac{{x}^{2} + 3}{{x}^{2} + 4}$ we can see that it is clear that the limit exists, and is approximately $1$

graph{(x^2+3)/(x^2+4) [-10, 10, -2, 2]}

Now, As $x \to - \infty$ then ${x}^{2} \to \infty$ ,but $\frac{1}{x} ^ 2 \to 0$

So, it would be better if we could replace ${x}^{2}$ with $\frac{1}{x} ^ 2$, or ${x}^{-} 2$

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

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

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

$\therefore {\lim}_{x \to - \infty} \frac{{x}^{2} + 3}{{x}^{2} + 4} = {\lim}_{x \to - \infty} \frac{1 + 3 {x}^{-} 2}{1 + 4 {x}^{-} 2}$

And, using ${x}^{-} 2 \to 0$ as $x \to - \infty$ we have;

${\lim}_{x \to - \infty} \frac{{x}^{2} + 3}{{x}^{2} + 4} = \frac{1 + 0}{1 + 0} = 1$

Which is completely consistent with the above graph.

Nov 2, 2016

${\lim}_{x \to - \infty} \frac{{x}^{2} + 3}{{x}^{2} + 4} = 1$

#### Explanation:

We can also use L'Hospital's Rule as the limit is of an indeterminate form $\frac{\infty}{\infty}$

So By L'Hospital's Rule;

${\lim}_{x \to - \infty} \frac{{x}^{2} + 3}{{x}^{2} + 4} = {\lim}_{x \to - \infty} \frac{\left({x}^{2} + 3\right) '}{\left({x}^{2} + 4\right) '}$
$\therefore {\lim}_{x \to - \infty} \frac{{x}^{2} + 3}{{x}^{2} + 4} = {\lim}_{x \to - \infty} \frac{2 x}{2 x}$
$\therefore {\lim}_{x \to - \infty} \frac{{x}^{2} + 3}{{x}^{2} + 4} = {\lim}_{x \to - \infty} \left(1\right)$
$\therefore {\lim}_{x \to - \infty} \frac{{x}^{2} + 3}{{x}^{2} + 4} = 1$