# $\mathop {\lim }\limits_{x \to {0^ + }} \left( {\frac{1}{x} - \frac{1}{{\sqrt x }}} \right)\;$?

$\mathop {\lim }\limits_{x \to {0^ + }} \left( {\frac{1}{x} - \frac{1}{{\sqrt x }}} \right) = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{1/\sqrt x - 1}}{{\sqrt x }} = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{ - \frac{1}{2}{x^{ - 3/2}}}}{{\frac{1}{2}{x^{ - 1/2}}}} = \mathop {\lim }\limits_{x \to {0^ + }} \left( { - \frac{1}{x}} \right) = - \infty$. However, the answer is $\infty$. Can you help me spot my error? Thanks!

• Thank you all for clearing that up for me! – drawar Nov 12 '12 at 16:03
• Informally, you can see that $1/x$ is going to infinity pretty fast, but $1/\sqrt x$ is slow and is slowing down in cancelling the $1/x$. Thus it looks that in the big picture, $1/x$ will win and ultimately the limit would go to $+ \infty$. – Sawarnik Feb 2 '14 at 8:48

l'Hôpital's rule was incorrectly applied to $\mathop {\lim }\limits_{x \to {0^ + }} \frac{{1/\sqrt x - 1}}{{\sqrt x }}$. The numerator goes to $+\infty$, while the denominator goes to $0$.
Both factors of $\frac{1}{\sqrt x}\left(\frac{1}{\sqrt x}-1\right)$ go to $+\infty$. Or consider $\frac{1}{x}(1-\sqrt x)$, where the second factor goes to $1$.
You've misapplied L'Hopital rule. Your numerator tends to $+\infty$ and your denominator tends to $0$ (from above). Thus, the quotient tends to $+\infty$.
In this problem you will evaluate the right hand limit of the function at $$x = 0$$, this means that we need to find the limiting value of the function $$\frac{1}{x}-\frac{1}{\sqrt{x}}$$ as the value of $$x$$ reaches $$0$$ from a value that is slightly greater than $$0$$ or in other words a number which is infinitesimally greater than $$0$$. This can be approached by putting $$x=0+h$$. Now since $$x\rightarrow 0^{+}$$ , we can be sure that $$h\rightarrow 0$$. Now the problem becomes: $$\lim_{h \rightarrow 0}(\frac{1}{h} -\frac{1}{\sqrt{h}})$$ or $$\lim_{h \rightarrow 0}(\frac{1-\sqrt{h}}{h})$$ Simply put $$h$$ as $$0$$ the limit becomes $$+\infty$$