How do you find lim (5+x^-1)/(1+2x^-1) as x->oo using l'Hospital's Rule or otherwise?

Jan 10, 2017

${\lim}_{x \rightarrow \infty} \frac{5 + {x}^{-} 1}{1 + 2 {x}^{-} 1} = 5$

Explanation:

In its present form it should be clear that you do not need to apply L'Hôpital's rule as ${x}^{-} 1 \rightarrow 0$ as $x \rightarrow \infty$ and so;

${\lim}_{x \rightarrow \infty} \frac{5 + {x}^{-} 1}{1 + 2 {x}^{-} 1} = \frac{5 + 0}{1 + 0} = 5$

If you do want to apply L'Hôpital's rule the we should multiply numerator and denominator both by $x$ to get;

${\lim}_{x \rightarrow \infty} \frac{5 + {x}^{-} 1}{1 + 2 {x}^{-} 1} = {\lim}_{x \rightarrow \infty} \frac{x}{x} \frac{5 + {x}^{-} 1}{1 + 2 {x}^{-} 1}$
$\text{ } = {\lim}_{x \rightarrow \infty} \frac{5 x + 1}{x + 2}$

We now have an indeterminate form of the type $\frac{\infty}{\infty}$ so we can apply L'Hôpital's rule:

${\lim}_{x \rightarrow a} f \frac{x}{g} \left(x\right) = {\lim}_{x \rightarrow a} \frac{f ' \left(x\right)}{g ' \left(x\right)}$

having satisfied ourselves that our limit meets L'Hôpital's criteria to get

${\lim}_{x \rightarrow \infty} \frac{5 + {x}^{-} 1}{1 + 2 {x}^{-} 1} = {\lim}_{x \rightarrow \infty} \frac{\frac{d}{\mathrm{dx}} \left(5 x + 1\right)}{\frac{d}{\mathrm{dx}} \left(x + 2\right)}$
$\text{ } = {\lim}_{x \rightarrow \infty} \frac{5}{1} = 5$, as above.