lim as x goes to 0 from the positive direction of sqrt(x)log(x) correct me if I'm wrong but the actual limit at x=o does not exist yes?
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$\displaystyle \lim_{x \to 0^{+}} \sqrt{x} \ln x = \lim_{x \to 0+} \frac{\ln x}{\frac{1}{\sqrt{x}}} = \lim_{x \to 0^{+}} \frac{\frac{1}{x}}{\frac{-1}{2x^{\frac{3}{2}}}} = \lim_{x \to 0^{+}} -2 \sqrt{x} = 0$
Thanks a heap, but does the limit at x=0 exist?
Does 0 exist?
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