Do limit exist for the following functions:
- $\lim_{x\rightarrow 0} \cos(\frac{1}{x})$
I think it exists because the expression for Left Hand Limit & Right Hand Limit are same
i.e $\lim_{h\rightarrow 0} \cos(\frac{1}{h})$ for $x=0+h$ & $x=0-h$
- $\lim_{x\rightarrow 0} \sin(\frac{1}{x})$
I think the limit doesn't exist because the expression for Left Hand Limit & Right Hand Limit are different.
$LHL:-\lim_{h\rightarrow 0} \sin(\frac{1}{h})$ for $x=0-h$
and
$RHL:\lim_{h\rightarrow 0} \sin(\frac{1}{h})$ for $x=0+h$
Is my thought process correct?
If the above is correct, then can we say the following with similar arguments:
(i)Limit exists for $RHL:\lim_{x\rightarrow 0} \dfrac{1}{| x |}$
(ii)Limit doesn't exist for $RHL:\lim_{x\rightarrow 0} \dfrac{1}{x}$
Best Answer
Actually, for both functions $\cos\left(\frac 1x\right)$ and $\sin \left(\frac 1x\right)$, the limits as $x\to 0$ do not exist, and for the same reasons, irregardless of whether we consider the limit as $x \to 0^+$ or $x\to 0^{-1}$.
Look, for example, of the behavior of $\cos \left(\frac 1x\right)$ on the interval $(-0.1, 0.1)$.
For any function $f(x)$, for a limit $L$ to exist as $x \to a$, we have $$\lim_{x\to a}f(x)=L\in \mathbb R\iff \forall(x_n)\to a,\; f(x_n)\to L$$
For both $f(x) = \cos \frac 1x$ and $f(x) = \sin \frac 1x$, there is no such $L$ to which $f(x)$ as $x \to 0$ from the right OR from the left.