$\mathop {\lim }\limits_{x \to {0^ + }} \left( {\frac{1}{x} - \frac{1}{{\sqrt x }}} \right)\;$? $\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!
 A: 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 $1$.
A: You've misapplied L'Hopital rule. Your numerator tends to $+\infty$ and your denominator tends to $0$ (from above). Thus, the quotient tends to $+\infty$.
A: 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 $+\infty$
A: $$\lim_{x\rightarrow 0^+}\left ( \frac{1}{x}-\frac{1}{\sqrt{x}} \right )=-\lim_{x\rightarrow 0^+}\frac{\sqrt{x}-1}{x}=-\lim_{x\rightarrow 0^+}\left ( \sqrt{x}-1 \right )\lim_{x\rightarrow 0^+}\frac{1}{x}=-1\lim_{x\rightarrow 0^+}\frac{1}{x}=-\left ( -\infty  \right )=\infty $$
