Evaluate: `("lim")_(xvec2)(int0"x"cost^2dt)/x`

Explanation

To evaluate the limit

\[ \lim_{x \to 0} \frac{\int_0^x \cos(t^2) \, dt}{x}, \]

we first observe that substituting \(x = 0\) gives us the indeterminate form \(\frac{0}{0}\). Therefore, we can apply L'Hôpital's Rule, which states that if we have a limit of the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), we can differentiate the numerator and the denominator separately.

### Step 1: Differentiate the Numerator and Denominator

The numerator is \(\int_0^x \cos(t^2) \, dt\) and the denominator is \(x\). We need to differentiate both:

- The derivative of the denominator \(x\) is \(1\).
- For the numerator, we can use the Fundamental Theorem of Calculus. The derivative of \(\int_0^x f(t) \, dt\) with respect to \(x\) is \(f(x)\). Hence, we have:

\[ \frac{d}{dx} \left( \int_0^x \cos(t^2) \, dt \right) = \cos(x^2). \]

### Step 2: Apply L'Hôpital's Rule

Now we can apply L'Hôpital's Rule:

\[ \lim_{x \to 0} \frac{\int_0^x \cos(t^2) \, dt}{x} = \lim_{x \to 0} \frac{\cos(x^2)}{1}. \]

### Step 3: Evaluate the Limit

Now we evaluate the limit:

\[ \lim_{x \to 0} \cos(x^2). \]

As \(x\) approaches \(0\), \(x^2\) also approaches \(0\). Therefore:

\[ \cos(0) = 1. \]

### Final Answer

Thus, the limit evaluates to:

\[ \lim_{x \to 0} \frac{\int_0^x \cos(t^2) \, dt}{x} = 1. \]

### Summary

The final answer is:

\[ \boxed{1}. \]