show function does not converge as h tends to 0

Question

I need to show that $3h\sin(1/h) - \cos(1/h)$ does not have a limit as $h \to 0$. I'm fairly new to the epsilon delta proofs and am not sure how to do it for this question, would someone be able to walk me through it please?

Answer 1

We have to prove $\lim_{x \to 0} (3hsin(\frac{1}{h}) - cos(\frac{1}{h}))$= Limit does not exist

So,

1).Lets prove $\lim_{x \to 0} 3hsin(\frac{1}{h})=0$

$$-1\leq sin\frac{1}{h}\leq 1$$

$$-3h\leq 3hsin\frac{1}{h}\leq 3h$$

$$0=\lim_{h \to 0}(-3h)\leq 3h\lim_{h \to 0}(sin\frac{1}{h})\leq \lim_{h \to 0}(3h)=0$$

By sandwich theorem we get, $$3h\lim_{h \to 0}sin\frac{1}{h}=0$$

2).Lets prove $\lim_{x \to 0} cos(\frac{1}{h})=Does \: not \: exist$

$$\forall \varepsilon >0 \; \exists \delta \; s.t \; 0<\left | x \right | < \delta \mapsto |cos\frac{1}{h}-L|<\varepsilon$$

Let $\varepsilon=\frac{1}{2}$,

$$0<\left | x \right | < \delta \mapsto |cos\frac{1}{h}-L|<\frac{1}{2}$$

So lets assume that, $$\lim_{x \to 0} cos(\frac{1}{h})=L \; \in \mathbb{R}$$

Let $$x_{1}=\frac{1}{2n\pi} \; \; (n\in Z^{+})\; \; and \; \; x_{2}=\frac{1}{2n\pi+\pi } \; \; (n\in Z^{+})$$ So we have to get range of n, $$0<|\frac{1}{2n\pi +\pi}| < |\frac{1}{2n\pi }|<\delta$$ $$0<\frac{1}{|2n\pi| }<\delta$$ So we get, $$0<\frac{1}{|2\pi\delta| }<n\; \; (n\in Z^{+})$$

Now, $$|cos(2n\pi)-L|=|cos(x_{1})-L|=|1-L|<\frac{1}{2}\Rightarrow \mathbf{A}$$ $$|cos(2n\pi+\pi)-L|=|cos(x_{2})-L|=|-1-L|=|1+L|<\frac{1}{2}\Rightarrow \mathbf{B}$$ By A + B we get, $$|1-L|+|1+L|< 1$$ $$|(1-L)+(1+L)|\leq |1-L|+|1+L|< 1$$ $$2\leq |1-L|+|1+L|< 1$$ $$2<1 \; \;\; \; (\therefore This \;is \;a \;contradiction)$$

So our assumption is false, Therefore $\lim_{x \to 0} cos(\frac{1}{h})$ has no limit.

So $\lim_{x \to 0} (3hsin(\frac{1}{h}) - cos(\frac{1}{h}))=\lim_{x \to 0} 3hsin(\frac{1}{h})-\lim_{x \to 0} cos(\frac{1}{h})=$Limit does not exist