Question

1. \\(\lim\limits_{x\to 1} \frac{\int_{1}^{x} \cos t \\, dt}{x^2 - 1}\\)

Explanation:

Step1: Apply L'Hospital's Rule

Since as \(x \to 1\), the numerator \(\int_{1}^{x}\cos t\ dt\) approaches \(0\) (because \(\int_{1}^{1}\cos t\ dt = 0\)) and the denominator \(x^{2}-1\) also approaches \(0\) (because \(1^{2}-1 = 0\)), we can use L'Hospital's Rule. L'Hospital's Rule states that if \(\lim_{x\to a}\frac{f(x)}{g(x)}\) is of the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then \(\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f^{\prime}(x)}{g^{\prime}(x)}\), provided the latter limit exists.

First, find the derivative of the numerator. By the Fundamental Theorem of Calculus, if \(F(x)=\int_{a}^{x}f(t)\ dt\), then \(F^{\prime}(x) = f(x)\). So, if \(f(t)=\cos t\) and the numerator is \(\int_{1}^{x}\cos t\ dt\), its derivative with respect to \(x\) is \(\cos x\).

Next, find the derivative of the denominator. The denominator is \(x^{2}-1\), and its derivative with respect to \(x\) is \(2x\).

So, by L'Hospital's Rule, \(\lim_{x\to 1}\frac{\int_{1}^{x}\cos t\ dt}{x^{2}-1}=\lim_{x\to 1}\frac{\cos x}{2x}\).

Step2: Evaluate the new limit

Now we can evaluate the limit \(\lim_{x\to 1}\frac{\cos x}{2x}\) by substituting \(x = 1\) into the function \(\frac{\cos x}{2x}\).

We know that \(\cos(1)\) is the cosine of \(1\) radian, and substituting \(x = 1\) into the denominator gives \(2\times1 = 2\). So, \(\frac{\cos(1)}{2\times1}=\frac{\cos 1}{2}\).

Answer:

\(\frac{\cos 1}{2}\)