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# Compute the limit of the given function

Evaluate the limit.

Both the numerator and denominator go to 0 as so we apply L’Hopital’s rule,

The initial limit can be transformed to $\lim_{x \to 0+} \frac{\sin x \log x}{\cos x}$. The denominator is 1, so let’s focus on the numerator. By application of Taylor’s expansion $\lim_{x \to 0+} (x + o(x)) \log x$. $\lim_{x \to 0+} x \log x = \lim_{t \to +\infty} \frac{-\log t}{t} = 0$ by T7.11. Similarly the limit for $o(x) \log x$ can be computed to be 0. Overall the numerator is 0, and the limit 0/1 = 0.