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Mathematics exercises
Let f : R ‒› R defined by f(x)=x^2.sin(1/x).
Show that f can be extended by continuity at 0;
f is denoted as an extended function.
Show that f is differentiable on R but that f' is not continuous at 0.
Here, the given function f is defined by f(x)= x^2.sin(1/x)
if x belongs to R,
so x is a real other than zero.
We want to extend f in a continuous function on R.
We know, thanks to the ordinary rules of operations on continuous functions,
that the function f is continuous on R.
But actually, we want to know if it can be extended by continuity at 0.
In other words, we are examining if f has a limit finishing at 0.
To do this, we use the formula of f(x), even the absolute value of f(x);
since it is an absolute value, it is positive or equal to 0.
By definition of f, it is equal to the absolute value of sin x^2.sin(1/x) - with x^2 positive -
and the absolute value of sin(1/x) is the sine of a real number:
we know that the absolute value is less than or equal to 1.
So, this quantity is less than or equal to x^2.
When x tends to 0, x^2 tends to 0.
So, according to the squeeze theorem,
f(x), which is between 0 and a quantity that tends to 0,
also tends to 0 (when x tends to 0).
Therefore, the function can be extended by continuity by setting f(0)=0
and we obtain a function that is continue to all of R.
Then, we need to investigate the differentiability of the function f.
So f is defined by f(x)=x^2.sin(1/x) if x is other than zero
and by 0 if x=0 according to the previous question.
To examine the differentiability, we start by examining the differentiability on R;
so here we have a product and a composite of functions that are differentiable on R.
Thus, the formula x^2.sin(1/x) is differentiable on R
and we even find the explicit formula of f'(x).
So f'(x) is the derivative of a product, which gives:
2x.sin(1/x) + x^2 multiplied by the derivative of sin(1/x).
Let me remind you the formula of the derivative of sinu: (sin u)'=u' cos u ;
here u=(1/x) so the derivative of sin(1/x)=(-1/x^2.cos(1/x)).
The formula is slightly reduced and we obtain f'(x)=2x.sin(1/x) – cos(1/x).
Now, we just need to investigate the case x=0.
In this case, it is not possible to apply the same rule
since f(0) is not defined by the formula x^2.sin(1/x) but by f(0)=0.
So we go back to the definition of the differentiability
and we look at the growth rate, which is f(x)-f(0)/x-0.
Here if we pay attention to the x other than 0,
the expression of f(x) is x^2.sin(1/x) ;
f(0) by definition of f is equal to 0.
The growth rate of f at 0 is therefore equal to x sin(1/x).
The same reasoning as in question 1 shows that this quantity tends to 0 when x tends to 0.
This shows that the function f is differentiable at 0 and that f'(0)=0.
Now, we are asked to see if the derivative f' is continue on R.
So f' (which we've just found) is that f'(x) is well defined
and is equal to: 2x.sin(1/x) – cos(1/x) if x is other than zero
(so if x belongs to R) and f'(0)=0.
Actually, we have to consider if this function is continuous on R.
It is continuous on R, still according to the ordinary rules of the products and composites of continuous functions.
The problem is to know if it is continuous at 0.
So f' is continuous at 0, which means f'(x) tends to f'(0)
when x tends to 0 by a value other than zero;
(it is given by this formula here).
We need to check if this quantity tends to 0 when x tends to 0.
Then, it is all right for the first term,
still with the same reasoning, we have x.1sin,
in other words, something which is between, -1 and 1;
so this quantity indeed tends to 0 when x tends to 0.
However, cos(1/x) has no limit when x tends to 0 since if x tends to 0,
1/x tends to infinity and the cosine has no limit at infinity.
Thus, something that tends to 0 minus something that has no limits
gives something that has no limits.
Consequently, this affirmation was wrong:
in other words, f' is not continuous at 0.
Here, the graph of the function f defined by f(x)=x^2.sin(1 /x) has been drawn.
We notice that – as the sine – it is an oscillating function.
But as the sine is between -1 and 1,
the function f(x) is between -x^2 et +x^2.
We can see very well on the drawing that f(x) tends to 0 when x tends to 0.
However, the derivative of the function f is not continuous at 0.
This is due to the oscillation of the function.
And when x tends to 0, (1/x) tends to infinity,
so the sine's oscillations go faster and faster
and the slope of the function is going to go indefinitely from positive values to negative values,
and it will go even faster when x tends to 0.
Authors: Arnaud Bodin, Léa Blanc-Centi, university of Lille 1/ Project Exo7.
Production/ Technical means: SEMM, Teaching and Multimedia service
University of Lille 1
UNISCIEL (The Online Thematic University for Sciences).