91Ó°ÊÓ

To analyze the behavior of function (A) \(\frac{\sin x \pi}{x}\) as x approaches infinity, we can remember L'Hopital's Rule which states that if the limit of the ratio of two functions is indeterminate at a certain point, then the limit of the ratio of their derivatives is equal to the original limit, i.e: $$\lim_{x\to\infty}\frac{f(x)}{g(x)} = \lim_{x\to\infty}\frac{f'(x)}{g'(x)}$$ So, applying L'Hopital's Rule, let's derive the numerator and the denominator: $$f(x) = \sin x\pi, f'(x) = \pi\cos x\pi$$ $$g(x) = x, g'(x) = 1$$ Now, let's find the limit: $$\lim_{x\to\infty} \frac{\sin x\pi}{x} = \lim_{x\to\infty} \frac{\pi\cos x\pi}{1}$$ Since the values of \(\cos x\pi\) oscillates between -1 and 1, and the value of \(\pi\) is constant, the limit does not exist for Function (A).

Let's analyze the limit of function (B): \(a \cos^2 x\pi + b \sin^2 x\pi\) as x approaches infinity. Notice that the function is composed of two trigonometric functions, and their values oscillate between [0, 1]. Therefore, no matter the values of a and b, the function will also oscillate as x approaches infinity. Thus, the limit does not exist for Function (B).

For function (C): \(x\sin x\pi\), let's analyze its behavior as x approaches infinity. Notice that the sine function's values oscillate between [-1, 1], and when multiplied by x, the resulting function will have a larger variation as x grows. Since x is unbounded, and sine's values keep oscillating, this function also does not have a limit as x approaches infinity.

Finally, let's analyze the behavior of function (D): \(\tan x\pi\) as x approaches infinity. We know that \(\tan x\) is undefined at odd multiples of \(\frac{\pi}{2}\), which causes the function to have vertical asymptotes at these points and thus making the function periodic. Therefore, as x approaches infinity, the function continues oscillating between the asymptotes, and the limit does not exist. In conclusion, each of the given functions does not have a limit as x approaches infinity.