Questions: Consider the function f(x)=sin x log x^3. It is true that Select one: lim x -> 0+ f(x) does not exist lim x -> 0+ f(x)=0 lim x -> 0+ f(x)=-∞ lim x -> 0+ f(x)=+∞ lim x -> 0+ f(x)=1

Transcript text: Consider the function $f(x)=\sin x \log x^{3}$. It is true that Select one: $\lim _{x \rightarrow 0^{+}} f(x)$ does not exist $\lim _{x \rightarrow 0^{+}} f(x)=0$ $\lim _{x \rightarrow 0^{+}} f(x)=-\infty$ $\lim _{x \rightarrow 0^{+}} f(x)=+\infty$ $\lim _{x \rightarrow 0^{+}} f(x)=1$

Solution

Solution Steps

Step 1: Define the Function

We start with the function $ f(x) = \sin x \log x^3 $. We need to evaluate the limit of this function as $ x $ approaches $ 0^+ $.

Step 2: Analyze the Components

As $ x $ approaches $ 0^+ $:

$ \sin x $ approaches $ 0 $.
$ \log x^3 = 3 \log x $ approaches $-\infty$.

Step 3: Identify the Indeterminate Form

The limit $ \lim_{x \to 0^+} f(x) $ results in the form $ 0 \cdot (-\infty) $, which is indeterminate. We need to resolve this indeterminate form.

Step 4: Rewrite the Function

We can rewrite the function as: \[ f(x) = \sin x \cdot 3 \log x = \frac{3 \log x}{\frac{1}{\sin x}} \] This transformation allows us to analyze the limit as a fraction.

Step 5: Apply L'Hôpital's Rule

Since we have the form $ \frac{-\infty}{\infty} $, we can apply L'Hôpital's Rule. We differentiate the numerator and the denominator:

The derivative of the numerator $ 3 \log x $ is $ \frac{3}{x} $.
The derivative of the denominator $ \frac{1}{\sin x} $ is $ -\frac{\cos x}{\sin^2 x} $.

Step 6: Evaluate the Limit

Now we evaluate the limit: \[ \lim_{x \to 0^+} \frac{3 \log x}{\frac{1}{\sin x}} = \lim_{x \to 0^+} \frac{3/x}{-\cos x/\sin^2 x} \] This simplifies to: \[ \lim_{x \to 0^+} -3 \frac{\sin^2 x}{x \cos x} \]

Step 7: Simplify Further

As $ x $ approaches $ 0^+ $: