Calculus and Differential Equations

4 marks: Key differences between limit and continuity

Let the point be $x=a$.

Relation to the graph

• Limit: If $\lim_{x\to a}f(x)=L$, then as $x$ approaches $a$ from either side, the graph approaches the height $L$. The point $(a,L)$ may be missing, or the actual value $f(a)$ may be different from $L$.
• Continuity: The graph passes through the point $(a,f(a))$ without a break. Locally, the curve can be drawn through $x=a$ without lifting the pencil.

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6 marks: Differentiability implies continuity, but not conversely

Theorem

If a function $f$ is differentiable at $x=a$, then $f$ is continuous at $x=a$.

Proof

Since $f$ is differentiable at $x=a$, the derivative

$$f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$$

exists and is finite.

For $h\ne 0$,

$$f(a+h)-f(a)=\frac{f(a+h)-f(a)}{h}\cdot h.$$

Taking the limit as $h\to 0$,

$$\lim_{h\to 0}\bigl[f(a+h)-f(a)\bigr] = \lim_{h\to 0}\frac{f(a+h)-f(a)}{h}\cdot \lim_{h\to 0}h.$$

Therefore,

$$\lim_{h\to 0}\bigl[f(a+h)-f(a)\bigr] = f'(a)\cdot 0 = 0.$$

Hence,

$$\lim_{h\to 0}f(a+h)=f(a).$$

Putting $x=a+h$, as $h\to 0$, we have $x\to a$. Thus,

$$\lim_{x\to a}f(x)=f(a).$$

So $f$ is continuous at $x=a$. Hence differentiability implies continuity. $\blacksquare$

Converse is not true

Continuity does not imply differentiability. For example, consider

$$f(x)=|x|$$

at $x=0$.

First, $f$ is continuous at $0$ because

$$\lim_{x\to 0}|x|=0=f(0).$$

But differentiability at $0$ requires

$$f'(0)=\lim_{h\to 0}\frac{|h|-0}{h} = \lim_{h\to 0}\frac{|h|}{h}.$$

Now,

$$\frac{|h|}{h}= \begin{cases} 1, & h>0,\\ -1, & h<0. \end{cases}$$

So the right-hand derivative is $1$ and the left-hand derivative is $-1$. Since these are not equal,

$$\lim_{h\to 0}\frac{|h|}{h}$$

does not exist. Therefore, $f(x)=|x|$ is continuous at $0$ but not differentiable at $0$.

The graph of $y=|x|$ has a sharp corner at $(0,0)$, so continuity holds but differentiability fails. Hence the converse is false. $\blacksquare$

(b) Homogeneous differential equation:
A first order and first degree differential equation of the form
[ \frac{dy}{dx}=f(x,y) ] is called a homogeneous differential equation if (f(x,y)) is a homogeneous function of degree zero, i.e.
[ f(tx,ty)=f(x,y). ] Equivalently, it can be written as
[ \frac{dy}{dx}=F\left(\frac yx\right) ] or as (M(x,y),dx+N(x,y),dy=0), where (M) and (N) are homogeneous functions of the same degree. Such equations are solved by putting (y=vx).

Now solve: [ \frac{dy}{dx}=\frac{x+y}{x-y} ]

The right-hand side is homogeneous of degree zero. Put
[ y=vx ] Then
[ \frac{dy}{dx}=v+x\frac{dv}{dx} ]

Substitute in the given equation: [ v+x\frac{dv}{dx}=\frac{x+vx}{x-vx} ] [ v+x\frac{dv}{dx}=\frac{1+v}{1-v} ]

So, [ x\frac{dv}{dx}=\frac{1+v}{1-v}-v ] [ x\frac{dv}{dx}=\frac{1+v-v+v^2}{1-v} ] [ x\frac{dv}{dx}=\frac{1+v^2}{1-v} ]

Separate the variables: [ \frac{1-v}{1+v^2},dv=\frac{dx}{x} ]

Integrating both sides: [ \int \frac{1}{1+v^2},dv-\int \frac{v}{1+v^2},dv=\int \frac{dx}{x} ] [ \tan^{-1}v-\frac12\ln(1+v^2)=\ln|x|+C ]

Now put (v=\frac yx): [ \tan^{-1}\left(\frac yx\right)-\frac12\ln\left(1+\frac{y^2}{x^2}\right)=\ln|x|+C ]

Simplifying, [ \tan^{-1}\left(\frac yx\right)-\frac12\ln(x^2+y^2)+\ln|x|=\ln|x|+C ] [ \tan^{-1}\left(\frac yx\right)-\frac12\ln(x^2+y^2)=C ]

Hence the general solution is
[ \boxed{\tan^{-1}\left(\frac yx\right)=\frac12\ln(x^2+y^2)+C} ] where (C) is an arbitrary constant.