How do you find the limit of (2-e^x)/(2+3e^x) as x approaches infinity?

Nov 1, 2016

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

Explanation:

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

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

Now, As $x \to \infty$ then ${e}^{x} \to \infty$ ,but ${e}^{-} x \to 0$

So, it would be better if we could replace ${e}^{x}$ with ${e}^{-} x$

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

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

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

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

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

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

Which is completely consistent with the above graph.