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

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

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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 $1$.

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Oops, I see my problem now. Thank you :) –  drawar Nov 12 '12 at 16:01

You've misapplied L'Hopital rule. Your numerator tends to $+\infty$ and your denominator tends to $0$ (from above). Thus, the quotient tends to $+\infty$.

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