Find lim as x to 0+ ln x tan x

Find $\displaystyle\lim_{x\to0^{+}}\ln x\,\tan x$.

Step 1: The form

As $x\to0^{+}$ we have $\ln x\to-\infty$ and $\tan x\to0$: the product is the indeterminate form $0\cdot\infty$. It must first be rewritten as a quotient before l'Hôpital's rule applies.

Step 2: Rewrite as a quotient

Divide by the reciprocal of the factor that tends to zero, $\tan x = 1/\cot x$: $\ln x\,\tan x = \frac{\ln x}{\cot x}$.

which as $x\to0^{+}$ is $\dfrac{-\infty}{+\infty}$ — a legitimate $\dfrac{\infty}{\infty}$.

Step 3: Apply l'Hôpital's rule

\[ \bigl(\ln x\bigr)' = \frac1x, \qquad \bigl(\cot x\bigr)' = -\frac{1}{\sin^{2}x}, \]

\[ \lim_{x\to0^{+}}\frac{\ln x}{\cot x} = \lim_{x\to0^{+}}\frac{1/x}{-1/\sin^{2}x} = -\lim_{x\to0^{+}}\frac{\sin^{2}x}{x}. \]

Step 4: Evaluate the new limit

Split off one standard factor: $\frac{\sin^{2}x}{x} = \sin x\cdot\frac{\sin x}{x} \longrightarrow 0\cdot1 = 0$.

Step 5: Verification

\[x = 0.1:\ -0.2310291; \qquad x = 0.01:\ -0.0460532; \qquad x = 0.001:\ -0.0069078 \;\;\checkmark\]

\[\tan x\sim x \;\Longrightarrow\; \ln x\,\tan x \sim x\ln x \to 0 \;\;\checkmark\]

\[\text{On } (0,1) \text{ the product dips to a minimum } -0.3875 \text{ at } x \approx 0.4100 \;\;\checkmark\]