Proving a limit using squeeze rule

Question

I'm trying to figure out how to calculate the limit of

$f(x)= \begin{cases} \frac{3}{2}\cdot x^{\frac{1}{2}}\cdot\sin\left(\frac{1}{x}\right)-\frac{1}{\sqrt{x}}\cdot\cos\left(\frac{1}{x}\right) & x > 0 \\ \\ 0 & x≤0 \\ \end{cases}$

as $x$ tends to $0^{+}$ I know it doesn't exist and tends to $-∞$, I've been trying forever to prove there exist two functions that are bigger/smaller than $f$ in order to use the squeeze rule but I'm begging to think there is some other easier way..

Answer 1

$\displaystyle \lim_{x\to 0^-} f(x)=0.$ But $\displaystyle \lim_{x\to 0^+} f(x)$ does not exist.

Use sequential criterion to prove that $\displaystyle \lim_{x\to 0} f(x)$ does not exist. Consider two sequences : $x_n={2\over (4n+1)\pi}$ and $x'_n={1\over 2n\pi}.$

See that: $\displaystyle \lim_{n\to \infty} f(x_n)=\lim_{n\to \infty}\textstyle \frac 32\cdot \frac{\sqrt 2}{\sqrt {(4n+1)\cdot \pi}}=0.$ But $\displaystyle \lim_{n\to \infty} f(x'_n)=\lim_{n\to \infty} -\sqrt{2n\pi}\textstyle \to -\infty$. Thus $\displaystyle \lim_{n\to \infty} x_n=\lim_{n\to \infty} x'_n=0$ but $\displaystyle \lim_{n\to \infty} f(x_n)\neq \lim_{n\to \infty} f(x'_n)$.

$\therefore \displaystyle \lim_{x\to 0} f(x)$ does not exist.

Answer 2

It is better to use a well known fact that as $x\to 0$ the functions $\sin(1/x),\cos (1/x)$ oscillate finitely between $1$ and $-1$. Most introductory calculus textbooks will present this example of oscillating function while dealing with the concept of limits.

Now as $x\to 0^+$ we can see that the first term of $f(x) $ namely $(3/2)x^{1/2}\sin(1/x)$ lies between $-(3/2)x^{1/2}$ and $(3/2)x^{1/2}$. Thus by Squeeze Theorem this term tends to $0$.

The next term $(1/\sqrt{x})\cos(1/x)$ consists of a factor which tends to $\infty $ and the other factor oscillates between $1$ and $-1$ and hence the term oscillates infinitely between $\infty$ and $-\infty $.

It follows that $f(x) $ is made of two terms one of which has limit $0$ and the other one oscillates infinitely. Hence $f(x) $ oscillates infinitely as $x\to 0^+$ and the limit of $f(x) $ as $x\to 0^+$ does not exist.