How to evaluate this limit with l'hopital's rule

Question

is it possible to use L'hopital for this or is there another method I'm missing? I have no idea how to even start this.

$$\lim_{x\to \infty} \frac{(9x+1)^\frac12}{x+1} $$

Answer 1

As far as I know, L'Hospital's rule can be applied, whenever you have a limit of a fraction and the numerator and denominator evaluated on their own and then put together give an undefined result like $\frac{0}{0}$ or $\frac{\infty}{\infty}$ and similar.

The rule then is (in it's simplest form, adjusted to this case):

$$\lim_{x\to\infty}\frac{f(x)}{g(x)}=\lim_{x\to\infty}\frac{f'(x)}{g'(x)}$$

In your case, we naively see something like $\frac{\infty}{\infty}$, right? So that with $f(x)$ and $g(x)$ as suggested by you we would get from L'Hospital:

$$\lim_{x\to\infty}\frac{(9x+1)^{\frac{1}{2}}}{x+1}=\lim_{x\to\infty}\frac{\frac{1}{2}(9x+1)^{-\frac{1}{2}}9}{1}=\lim_{x\to\infty}\frac{\frac{1}{2}9}{(9x+1)^{\frac{1}{2}}}=0$$

Answer 2

We can avoid L'Hospital's rule

Setting $\displaystyle \frac1x=h$

$$\lim_{x\to\infty^+}\frac{\sqrt{9x+1}}{x+1}=\lim_{h\to0^+}\frac{\sqrt h\sqrt{9+h}}{1+h}=\cdots$$

Answer 3

Hint $\ $ You asked for other methods. Here is one simple way.

$$ \dfrac{(9x+1)^{1/2}}{x+1\ \ \ } = \left(\dfrac{9x+1}{x^2\!+2x+1}\right)^{1/2} = \left(\dfrac{\dfrac{9}x +\dfrac{1}{x^2}}{1+\dfrac{2}x+\dfrac{1}{x^2}}\right)^{1/2}\ \ {\rm for}\ \ x> 0$$

Clearly, as $\,x\to\infty,\,$ the RHS has numerator $\to 0$ and denominator $\to 1$.