91Ó°ÊÓ

In this case, we are asked to find the limit as x approaches 1. So we first need to consider how the sine and cosine functions behave as x approaches 1. This will help us to determine the behavior of the functions in the given limit expression. \(\lim _{x \rightarrow 1} \frac{1+\sin \pi\left(\frac{3x}{1+x^{2}}\right)}{1+\cos \pi x}\) As x approaches 1, we can see that the arguments of the sine and cosine functions simplify as follows: \(\left(\frac{3(1)}{1+1^{2}}\right)=\frac{3}{2}\) And for cosine: \((\pi(1))=\pi\) Now, we know that \(\sin(\frac{3\pi}{2}) = -1\) and \(\cos(\pi) = -1\). We can substitute these values into the limit expression.

Substitute the values we found in the previous step into the limit expression: \(\lim _{x \rightarrow 1} \frac{1+\sin \pi\left(\frac{3x}{1+x^{2}}\right)}{1+\cos \pi x} = \frac{1+\sin \left(\frac{3\pi}{2}\right)}{1+\cos \pi} = \frac{1 - 1}{1 - 1}\)

Now, we can evaluate the limit by dividing the numerator and denominator: \(\frac{1 - 1}{1 - 1} = \frac{0}{0}\) The limit has the indeterminate form \(\frac{0}{0}\), which means that we need to utilize another method to find the limit.

Since the limit has the indeterminate form \(\frac{0}{0}\), we can use L'hopital's Rule, which states that if the limit of the ratio of the derivatives exists as x approaches a, then the limit of the ratio of the original functions also exists and is equal to the limit of the ratio of the derivatives: \(\lim _{x \rightarrow a} \frac{f(x)}{g(x)} = \lim _{x \rightarrow a} \frac{f'(x)}{g'(x)}\) Let \(f(x) = 1+\sin \pi\left(\frac{3x}{1+x^{2}}\right)\) and \(g(x) = 1+\cos \pi x\). We need to find the derivatives of these functions with respect to x, and then we can apply L'hopital's Rule.

Using the chain rule, we find the derivatives of f(x) and g(x) with respect to x. \(f'(x) = \pi \cos \pi\left(\frac{3x}{1+x^{2}}\right) \cdot \left(\frac{3(1+x^{2}) - 3x^{2}(2x)}{(1+x^{2})^{2}}\right) = \pi \cos \pi\left(\frac{3x}{1+x^{2}}\right) \cdot \left(\frac{3}{(1+x^{2})^{2}}\right)\) \(g'(x) = - \pi \sin \pi x\)

Now we apply L'hopital's Rule: \(\lim _{x \rightarrow 1} \frac{1+\sin \pi\left(\frac{3x}{1+x^{2}}\right)}{1+\cos \pi x} = \lim _{x \rightarrow 1} \frac{\pi \cos \pi\left(\frac{3x}{1+x^{2}}\right) \cdot \left(\frac{3}{(1+x^{2})^{2}}\right)}{- \pi \sin \pi x}\) As x approaches 1, the arguments of the cosine and sine functions simplify as before: \(\lim _{x \rightarrow 1} \frac{\pi \cos \left(\frac{3\pi}{2}\right) \cdot \left(\frac{3}{(1+1^{2})^{2}}\right)}{- \pi \sin (\pi)} = \frac{\pi(0) \cdot \left(\frac{3}{2^{2}}\right)}{- \pi(0)} = \frac{0}{0}\) Once again, we have arrived at the indeterminate form \(\frac{0}{0}\), which means that we cannot use L'hopital's Rule again, and we did something wrong in the previous steps. In this case, the correct answer is: (D) None of these