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

##### 2 Answers
Nov 3, 2016

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

#### Explanation:

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

graph{(e^x-x^2)/(e^x+x) [-5, 15, -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 ${e}^{-} x$

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

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

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

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

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

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

Which is completely consistent with the above graph.

Nov 3, 2016

1

#### Explanation:

Use iteratively L'Hospital's rule, until the indeterminate form $\frac{\infty}{\infty}$

disappears..

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

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

$= \lim x \to \infty \left(\frac{\left({e}^{x} - 2\right) '}{\left({e}^{x}\right) '}\right)$

$\lim x \to \infty \left(1 - 2 {e}^{- x}\right)$

$= \left(1 - 0\right) = 1$