A Rather Limited Integral

Similar solution as @John Rozmarynowycz 's

$\begin{aligned} I & = \int \left(2x \sin \frac 1x - \cos \frac 1x \right) dx \\ & = {\color{#3D99F6} \int 2x \sin \frac 1x \ dx} - \int \cos \frac 1x \ dx & \small \color{#3D99F6} \text{By integration by parts} \\ & = {\color{#3D99F6} x^2 \sin \frac 1x + C - \int x^2 \left(- \frac 1{x^2} \cos \frac 1x \right) dx} - \int \cos \frac 1x \ dx & \small \color{#3D99F6} \text{where }C \text{ is the constant of integration} \\ & = x^2 \sin \frac 1x + C + \int \cos \frac 1x \ dx - \int \cos \frac 1x \ dx \\ & = x^2 \sin \frac 1x + C \end{aligned}$

Therefore,

$\begin{aligned} \lim_{x \to 0} I \ \bigg|_{C=0} & = \lim_{x \to 0}x^2 \color{#3D99F6} \sin \frac 1x & \small \color{#3D99F6} \text{Note that }\lim_{x \to 0} \sin \frac 1x \in [-1, 1] \text{ is bounded.} \\ & = \boxed 0 \end{aligned}$

$\lim_{x\to 0} \Bigg[\int 2x\sin\bigg(\frac{1}{x}\bigg)-\cos\bigg(\frac{1}{x}\bigg)dx\Bigg]$

$\lim_{x\to 0} \Bigg[\int 2x\sin\bigg(\frac{1}{x}\bigg)dx - \int\cos\bigg(\frac{1}{x}\bigg)dx\Bigg]$

$\int f(x)g'(x)dx=f(x)g(x)-\int f'(x)g(x)dx$

$f(x)=\sin\bigg(\frac{1}{x}\bigg) \;\text{and}\; g'(x)= 2x$

then $f'(x)=-\frac{\cos\big(\frac{1}{x}\big)}{x^{2}} \;\text{and}\; g(x)=x^{2}$

$\int 2x\sin\bigg(\frac{1}{x}\bigg)dx=x^{2}\sin\bigg(\frac{1}{x}\bigg)+C-\int-\frac{\cos\big(\frac{1}{x}\big)}{x^{2}} x^{2}dx$

$=x^{2}\sin\bigg(\frac{1}{x}\bigg)+C+\int\cos\bigg(\frac{1}{x}\bigg)dx$

implement this expression into the integral and simplify

$=\lim_{x\to 0} \Bigg[x^{2}\sin\bigg(\frac{1}{x}\bigg)+C+\int\cos\bigg(\frac{1}{x}\bigg)dx-\int\cos\bigg(\frac{1}{x}\bigg)dx\Bigg]$

$=\lim_{x\to 0} \Bigg[x^{2}\sin\bigg(\frac{1}{x}\bigg)\Bigg]+C$

the range of values for the $\sin\big(\frac{1}{x}\big)$ function exists in the interval [-1,1]

$-1\leq \sin\bigg(\frac{1}{x}\bigg) \leq 1$

multiply the inequality by $x^{2}$ . All of the values that come from $x^{2}$ will be greater than or equal to zero, so it is unnecessary to flip any inequality signs.

$-x^{2}\leq x^{2}\sin\bigg(\frac{1}{x}\bigg) \leq x^{2}$ Next,

$\lim_{x\to 0} -x^{2}\leq \lim_{x\to 0} x^{2}\sin\bigg(\frac{1}{x}\bigg) \leq \lim_{x\to 0}x^{2}$

$0\leq\lim_{x\to 0} x^{2}\sin\bigg(\frac{1}{x}\bigg)\leq0$

therefore, by use of the squeeze theorem, it can be said that the limit of $x^{2}\sin\big(\frac{1}{x}\big)=0$

As such, it seems safe to assume that the limit of this integral is equal to zero, or,

$\lim_{x\to 0} \Bigg[\int 2x\sin\bigg(\frac{1}{x}\bigg)-\cos\bigg(\frac{1}{x}\bigg)dx\Bigg]=0$