Spivak Calculus Chapter 5-33d: $\lim_{x \to\infty}\frac{x^2(1+\sin^2(x))}{(x+\sin(x))^2}$ Find the following limit:
33d) $$\lim_{x\to\infty}\frac{x^2(1+\sin^2(x))}{(x+\sin(x))^2}$$
I am 99.9% sure this limit doesn't exist because as $x \rightarrow \infty$, $1+\sin^2(x)$ periodically oscillates between 1 and 2, which means so does the limit.
Is this a trick question or am I losing my mind?
 A: In the type of questions like the one you have mentioned the following methodology can be useful:
Assume we are faced with a complex expression:
$$\lim_{x\to\infty}\frac{f_1(x)f_2(x)}{g_1(x)g_2(x)}$$
In the case, the limit is of the form "$\frac{\infty}{\infty}$" the "Go to" attack will be to first identify the "dominant term" in the numerator and divide by it both the denominator and the numerator.
Assuming in our example $f_1(x)\geq f_2(x) \space\forall x>x_0$ we should try to analyse $$\lim_{x\to\infty}\frac{f_2(x)}{g_1(x)g_2(x)\cdot f_1(x)^{-1}}$$
In the particular case, we identify $x^2$ as the main term and divide by it, which yields:
$$\lim_{x\to\infty}\frac{1+\sin^2x}{1+2/x\cdot \sin x +1/x^2 \cdot \sin^2x}$$
Heuristically, now we see why there is no limit when x tends to infinity, the denominator is essentially $1$; however, the numerator oscillates between $[1,2]$.
To prove so rigorously we use Heine definition of the limit:

$\lim_{x\to\infty}f(x)=L$ if and only if  $\forall a_n\to_{n\to\infty}\infty$ $\lim_{n\to\infty}f(a_n)=L$

Hence, disproving will equate to producing two series $a_n,b_n$ such that $\lim_{n\to\infty}f(a_n) \neq \lim_{n\to\infty}f(b_n)$.
For example, we choose the extrema of $\sin$ $$a_n=2\pi n + \pi/2\space,\space b_n=2\pi n $$
Now evaluating the function on both:
$$\lim_{n\to\infty}f(a_n) = \lim_{n\to\infty}\frac{1+\sin^2(a_n)}{1+2/a_n\cdot \sin a_n +1/x^2 \cdot \sin^2a_n}=2$$
$$\lim_{n\to\infty}f(b_n) = \lim_{n\to\infty}\frac{1+\sin^2(b_n)}{1+2/b_n\cdot \sin b_n +1/x^2 \cdot \sin^2b_n}=1$$
Thus no limit exists.
