Why does $\lim_{k \rightarrow \infty} \lvert k^2 \sin (k^4) \rvert = 0$? For context: the question was if $x^2 y^2 \sin(\frac{1}{x^4}) + 2$ converges on $2$ for $x \rightarrow 0$. My initial assumption was that this is false because if I use the sequence $(x_k, y_k) = (\frac{1}{k}, k^2)$ this works out as $\mathop{\lim}\limits_{k \rightarrow \infty} \lvert k^2 \sin(k^4) \rvert \overset{?}{=} 0$, which intuitively seems to diverge.
Yet the correct answer to the question was that it does indeed converge, and Wolfram Alpha also says so. But plotting the sequence with GeoGebra for the first 1000 elements shows a somewhat "random", yet growing sequence1.
So why does $\lvert k^2 \sin(k^4) \rvert$ sequence converge?

1 Using the command Sequence[(k, abs(k² sin(k⁴))), k, 1, 1000]
 A: The limit in the title doesn't converge. The function $f(x) = x^2 \sin (x^4)$ has $f(x_n) \to \infty$ for the sequence $x_n = \left(\frac{\pi}{2} + 2\pi n\right)^{1/4}$ and $f(y_n) \to 0$ for the sequence $y_n =\left(2\pi n\right)^{1/4}$.
A: In multi-variable calculus, whenever when we study what happens with a function in multiple variables when we operate (i.e. derivate, integrate, take a limit) with one variable, it is (usually) understood that all the other variables are fixed and should be treated as constants. This is the standard convention, and most likely it was explained/stated in your course. 
If this usual convention was used in your class, then the question asks "If we fix $y$, then does $x^2 y^2 \sin(\frac{1}{x^4}) + 2$ converges to $2$ when $x \to 0$?". And the answer is yes."
Extra comment
Also note that if you calculated at any point an integral of the form
$$\int_0^1 xy dx $$
you actually used the convention that $y$ must be fixed. And if you set the corresponding Riemann sum for this integral, you end up with a limit of the form 
$$\lim_{n} \frac{1}{n} \sum_{k=1}^n \frac{k}{n}y$$
which exactly as in your example does not exists if you take $y=n \to \infty$, yet you know how to calculate integrals of this type. Same argument applies for derivatives.
This convention for integrals and derivatives (i.e. derivate/integrate wrt one variable and treat the rest as constants) is actually equivalent with this convention for limits: to calculate $\lim_x f(x,y)$ fix $y$ and calculate the limit by $x$. 
A: I am assuming (albeit without authority) that when Wolfram says $\lim_{k\rightarrow \infty}f(k) = 0$ to $\infty$, that means:
For any $N \in \mathbb R$ and any $\epsilon > 0$ and any $a \in [0,\infty)$ there is a $k > N$ so that $|f(k) - a| < \epsilon$.
i.e. $f(x)$ not only doesn't converge to any real or extended real value, it also oscillates between all non-negative limits.
For $f(k) = k^2\sin k^4$ this is clearly true.  $\sin k^4$ oscilates between all values between $0$ and $1$ and so $f(k)$ which is continuous will oscilate between all values between $0$ and all  $x < \lim k^2 = \infty$.
