# differentiability at point

• Mar 25th 2013, 07:50 AM
learning
differentiability at point
consider the function: $\displaystyle f(x) = x^2cos(1/x)$ if x $\displaystyle \neq 0$. 0 if x=0. show f is differentiable at 0? is f continuous at 0? if f'(x) continuous at 0?. ... ok that is the problem. but i dont know how to take the limit since it is cos(1/h) and then you cant just stick in the zero. i cant use l'hopital which is what i usually try. using the limit of differentiability i get the derivative is hcos(1/h) as h goes to 0. but i cant do anything with it.
• Mar 25th 2013, 11:31 AM
emakarov
Re: differentiability at point
Quote:

Originally Posted by learning
using the limit of differentiability i get the derivative is hcos(1/h) as h goes to 0. but i cant do anything with it.

The limit of differentiability, eh? So, differentiability is a function of which you can take a limit, right? Do you know the value of differentiability at 5?

Concerning the limit of $\displaystyle h\cos(1/h)$ as $\displaystyle h\to0$, what happens to a magnitude that is locked between -1 and 1 as you multiply it by smaller and smaller values?
• Mar 25th 2013, 01:22 PM
learning
Re: differentiability at point
come on now you ok i did mistype. sorry. i mean limit definition of differentiability (at a point/general. at a point here). as for your remark on this limit.. i know the limit is zero and i know why. but how can i state it to be so rather just by intuition? i get no mark if i just say it is obvious. ok so since the limit of h as h goes to zero, you are multiplying a 0 by .... well anything would make it zero. BUT, the limit of cos(1/h) as h goes to zero doesnt exist so i cant just use product of limits = limit of products.
• Mar 25th 2013, 01:38 PM
Plato
Re: differentiability at point
Quote:

Originally Posted by learning
come on now you ok i did mistype. sorry. i mean limit definition of differentiability (at a point/general. at a point here). as for your remark on this limit.. i know the limit is zero and i know why. but how can i state it to be so rather just by intuition? i get no mark if i just say it is obvious. ok so since the limit of h as h goes to zero, you are multiplying a 0 by .... well anything would make it zero. BUT, the limit of cos(1/h) as h goes to zero doesnt exist so i cant just use product of limits = limit of products.

$\displaystyle f'(a)={\lim _{h \to 0}}\frac{{f(a + h) - f(a)}}{h}$ and $\displaystyle f(x) = \left\{ {\begin{array}{*{20}{c}}{x^2\cos \left( {\frac{1}{x}} \right),}&{x \ne 0}\\{0,}&{x = 0}\end{array}} \right.$

$\displaystyle h\ne 0,~\frac{{f(0+ h) - f(0)}}{h}=h\cos \left( {\frac{1}{h} \right)$

$\displaystyle -|h|\le\left|h\cos \left( {\frac{1}{h} \right)\right|\le |h|$

Let $\displaystyle h\to 0$
• Mar 25th 2013, 01:56 PM
emakarov
Re: differentiability at point
If you have an intuitive understanding why $\displaystyle h\cos(1/h)$ tends to 0 (as a product of a function that tends to 0 and a bounded function), it is easy to write an epsilon-delta proof that the limit is zero. Let's proof a generalization.

Theorem. Let f(x) be such that $\displaystyle f(x)\to0$ as $\displaystyle x\to x_0$ and let g(x) be bounded in some neighborhood of $\displaystyle x_0$. The latter means that there exist positive constants r and M such that $\displaystyle |x - x_0| < r$ implies $\displaystyle |g(x)|\le M$ for all x. Then $\displaystyle f(x)g(x)\to0$ as $\displaystyle x\to0$.

Proof. By definition of limit, for every $\displaystyle \varepsilon>0$ there exist a $\displaystyle \delta>0$ such that $\displaystyle 0<|x-x_0|<\delta$ implies $\displaystyle |f(x)|<\varepsilon$ for all x. Suppose some $\displaystyle \varepsilon>0$ is given. Fix the corresponding $\displaystyle \delta$. What happens to $\displaystyle |f(x)g(x)|=|f(x)|\cdot|g(x)|$ when both $\displaystyle 0<|x-x_0|<\delta$ and $\displaystyle |x-x_0|<r$, i.e., when $\displaystyle 0<|x-x_0|<\min(\delta,r)$? Then $\displaystyle |f(x)g(x)|<M\varepsilon$. We would like to make $\displaystyle |f(x)g(x)|<\varepsilon$ to conform to the definition of limit. For this, we choose the $\displaystyle \delta$ that corresponds not to $\displaystyle \varepsilon$ but to $\displaystyle \varepsilon/M$. Then $\displaystyle 0<|x-x_0|<\delta$ implies $\displaystyle |f(x)|<\varepsilon/M$ and $\displaystyle 0<|x-x_0|<\min(\delta,r)$ implies $\displaystyle |f(x)g(x)|<\varepsilon$ for all x.

Altogether, given an $\displaystyle \varepsilon>0$, there exists a $\displaystyle \delta'$, namely, $\displaystyle \delta'=\min(\delta,r)$ where $\displaystyle \delta$ is obtained from the definition of limit of f(x) for $\displaystyle \varepsilon/M$. For this $\displaystyle \delta'$ it is true that $\displaystyle 0<|x-x_0|<\delta'$ implies $\displaystyle |f(x)g(x)|<\varepsilon$ for all x.

You are left with an exercise to understand this proof, to apply it to the current problem and simplify where possible for concrete $\displaystyle f(x) = x$ and $\displaystyle g(x) = \cos(1/x)$.