Evaluate `lim_(xto0)|x|^(sinx)`

Answer

To evaluate the limit \( \lim_{x \to 0} |x|^{\sin x} \), we can follow these steps:

### Step 1: Rewrite the expression using logarithms
We know that \( a^b = e^{b \ln a} \). Therefore, we can rewrite the limit as:
\[ |x|^{\sin x} = e^{\sin x \ln |x|} \]
Thus, we have:
\[ \lim_{x \to 0} |x|^{\sin x} = \lim_{x \to 0} e^{\sin x \ln |x|} \]

### Step 2: Analyze the limit of the exponent
Next, we need to evaluate the limit of the exponent:
\[ \lim_{x \to 0} \sin x \ln |x| \]
As \( x \to 0 \), \( \sin x \to 0 \) and \( \ln |x| \to -\infty \). This creates an indeterminate form of \( 0 \cdot (-\infty) \). To resolve this, we can rewrite it as:
\[ \sin x \ln |x| = \frac{\ln |x|}{\frac{1}{\sin x}} \]
Now, as \( x \to 0 \), \( \frac{1}{\sin x} \to \infty \) and \( \ln |x| \to -\infty \), giving us the form \( \frac{-\infty}{\infty} \).

### Step 3: Apply L'Hôpital's Rule
We can apply L'Hôpital's Rule to evaluate this limit:
\[ \lim_{x \to 0} \frac{\ln |x|}{\frac{1}{\sin x}} \]
Differentiating the numerator and denominator gives:
- The derivative of \( \ln |x| \) is \( \frac{1}{x} \).
- The derivative of \( \frac{1}{\sin x} \) is \( -\frac{\cos x}{\sin^2 x} \).

Thus, we have:
\[ \lim_{x \to 0} \frac{\frac{1}{x}}{-\frac{\cos x}{\sin^2 x}} = \lim_{x \to 0} -\frac{\sin^2 x}{x \cos x} \]

### Step 4: Simplify the limit
Now we can simplify the limit:
\[ \lim_{x \to 0} -\frac{\sin^2 x}{x \cos x} = \lim_{x \to 0} -\frac{\sin^2 x}{x^2} \cdot \frac{x}{\cos x} \]
We know that \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \), so:
\[ \lim_{x \to 0} -\frac{\sin^2 x}{x^2} = -1 \]
And since \( \cos x \to 1 \) as \( x \to 0 \):
\[ \lim_{x \to 0} \frac{x}{\cos x} = 0 \]
Thus, the limit evaluates to:
\[ \lim_{x \to 0} -\frac{\sin^2 x}{x \cos x} = 0 \]

### Step 5: Conclude the limit
Now substituting back into our expression for the limit:
\[ \lim_{x \to 0} e^{\sin x \ln |x|} = e^{0} = 1 \]

### Final Answer
Therefore, the limit is:
\[ \lim_{x \to 0} |x|^{\sin x} = 1 \]