stylistic:style_transfer
First example

The limit

lim x → 0 x 2 sin (1 x) {\displaystyle \lim _{x\to 0}x^{2}\sin \left({\tfrac {1}{x}}\right)}

cannot be determined through the limit law

lim x → a (f (x) ⋅ g (x)) = lim x → a f (x) ⋅ lim x → a g (x), {\displaystyle \lim _{x\to a}(f(x)\cdot g(x))=\lim _{x\to a}f(x)\cdot \lim _{x\to a}g(x),}

because

lim x → 0 sin (1 x) {\displaystyle \lim _{x\to 0}\sin \left({\tfrac {1}{x}}\right)}

does not exist.

However, by the definition of the sine function,

− 1 ≤ sin (1 x) ≤ 1. {\displaystyle -1\leq \sin \left({\tfrac {1}{x}}\right)\leq 1.}

It follows that

− x 2 ≤ x 2 sin (1 x) ≤ x 2 {\displaystyle -x^{2}\leq x^{2}\sin \left({\tfrac {1}{x}}\right)\leq x^{2}}

Since lim x → 0 − x 2 = lim x → 0 x 2 = 0 {\displaystyle \lim _{x\to 0}-x^{2}=\lim _{x\to 0}x^{2}=0}, by the squeeze theorem, lim x → 0 x 2 sin (1 x) {\displaystyle \lim _{x\to 0}x^{2}\sin \left({\tfrac {1}{x}}\right)} must also be 0.