First, we can condense the terms being added as
[tex]\cos\dfrac1{x^2}+\sin\dfrac1{x^2}=\sqrt2\sin\left(\dfrac1{x^2}+\dfrac\pi4\right)[/tex]
because [tex]\cos x[/tex] and [tex]\sin x[/tex] differ in phase by [tex]\dfrac\pi4[/tex]. Now using the fact that [tex]|\sin x|\le1[/tex], we have
[tex]-\sin x\le\sin x\left(\cos\dfrac1{x^2}+\sin\dfrac1{x^2}\right)\le\sin x[/tex]
so that by the squeeze theorem,
[tex]\displaystyle\lim_{x\to0}(-\sin x)\le\lim_{x\to0}\sin x\left(\cos\dfrac1{x^2}+\sin\dfrac1{x^2}\right)\le\lim_{x\to0}\sin x[/tex]
[tex]\displaystyle0\le\lim_{x\to0}\sin x\left(\cos\dfrac1{x^2}+\sin\dfrac1{x^2}\right)\le0[/tex]
and so the limit is 0.
