Does a limit exist at a cusp or sharp point

Question

Does a limit exist at a sharp curve or point? The limit as $x \to c+$ and $x \to c-$ won't equal at a cusp right? (or would it in some cases? So depending on that, does a limit always or never or sometimes exists at a cusp.

Thanks guys

Answer 1

Yes there exists a limit at a sharp point.

According to the definition of limit. Limit $L$ exists if $$\lim _{x\rightarrow n^+}f\left( x\right) =\lim _{x\rightarrow n^-}f\left( x\right) $$ The function is of course still continuous at the cusp so the limit exists and is evaluated as $$\lim _{x\rightarrow n^+}f\left( x\right) = L =\lim _{x\rightarrow n^-}f\left( x\right) $$

Answer 2

I asked myself this question as well and I tried to answer it based on the following example f(x)=|x| at the point x=0.

a) Does the limit exist at x=0?

b) Does the derivative exist at x=0?

One definition of a convergent functions is;

Let $f:D\to R$ be a function and a a cluster point of D. The function converges in a, if for $\forall a_n\in D$\{a}, which converges to a $f(a_n)$ converges as well. Based on the definition we can evaluate both questions;

a) Let $a_n=\frac{1}{n}$ be the sequence approaching 0 from the right and

$a_n=-\frac{1}{n}$ be the sequence approaching 0 from the left, then;

$\lim_{x\to0^+}f(x)=\lim_{n\to\infty^+}f(a_n)=\lim_{n\to\infty^+}|+\frac{1}{n}|=0$ $\lim_{x\to0^-}f(x)=\lim_{n\to\infty^+}f(a_n)=\lim_{n\to\infty^+}|-\frac{1}{n}|=0$

Thus the limit exists since $\lim_{x\to0^-}f(x)$=$\lim_{x\to0^+}f(x)$

b)

$\lim_{h\to0^+}\frac{f(a+h)-f(a)}{h}=\lim_{n\to\infty^+}\frac{f(a+a_n)-f(a)}{a_n}=\lim_{n\to\infty^+}\frac{f(0+\frac{1}{n})-f(0)}{\frac{1}{n}}=\lim_{n\to\infty^+}\frac{|0+\frac{1}{n}|-|0|}{\frac{1}{n}}=1$

$\lim_{h\to0^-}\frac{f(a+a_n)-f(a)}{a_n}=\lim_{n\to\infty^+}\frac{f(a-\frac{1}{n})-f(a)}{-\frac{1}{n}}=\lim_{n\to\infty^+}\frac{f(0-\frac{1}{n})-f(0)}{-\frac{1}{n}}=\lim_{n\to\infty^+}\frac{|0-\frac{1}{n}|-|0|}{-\frac{1}{n}}=\lim_{n\to\infty^+}\frac{\frac{1}{n}}{-\frac{1}{n}}=-1$

The limits are not equal and hence the derivative of the function at x=0 does not exist.

The above is not a proof but only an example. In general the limit exists for all x in the interior of the domain. The derivative however does not necessarily exist for all continuous functions.