# How do you find the limit (1+5/sqrtx)/(2+1/sqrtx) as x->0^+?

Nov 4, 2016

${\lim}_{x \to {0}^{+}} \left(\frac{1 + \frac{5}{\sqrt{x}}}{2 + \frac{1}{\sqrt{x}}}\right) = 5$

#### Explanation:

If we look at the graph of $y = \frac{1 + \frac{5}{\sqrt{x}}}{2 + \frac{1}{\sqrt{x}}}$ we can see that it is clear that the limit exists, and is approximately $5$

graph{(1+5/sqrt(x))/(2+1/sqrt(x)) [-8, 8, -2, 10]}

Now, As $x \to 0$ then $\frac{1}{x} \to \infty$ and $\frac{1}{\sqrt{x}} \to \infty$ but if we can invert these expressions they both $\to 0$

So, we look for a way to invert the $\frac{1}{A}$ expression

${\lim}_{x \to {0}^{+}} \left(\frac{1 + \frac{5}{\sqrt{x}}}{2 + \frac{1}{\sqrt{x}}}\right) = {\lim}_{x \to {0}^{+}} \frac{\sqrt{x}}{\sqrt{x}} \cdot \frac{1 + \frac{5}{\sqrt{x}}}{2 + \frac{1}{\sqrt{x}}}$

$\therefore {\lim}_{x \to {0}^{+}} \left(\frac{1 + \frac{5}{\sqrt{x}}}{2 + \frac{1}{\sqrt{x}}}\right) = \frac{\left({\lim}_{x \to {0}^{+}} \sqrt{x}\right) + 5}{\left({\lim}_{x \to {0}^{+}} \sqrt{x}\right) + 1}$

$\therefore {\lim}_{x \to {0}^{+}} \left(\frac{1 + \frac{5}{\sqrt{x}}}{2 + \frac{1}{\sqrt{x}}}\right) = \frac{0 + 5}{0 + 1}$

$\therefore {\lim}_{x \to {0}^{+}} \left(\frac{1 + \frac{5}{\sqrt{x}}}{2 + \frac{1}{\sqrt{x}}}\right) = 5$

Which is completely consistent with the above graph.