Differentiability and Continuity: Let f(x) be defined by x*sin(1/x) if x is not equal to 0 and by...

Question:

Differentiability and Continuity:

Let {eq}f(x) {/eq} be defined by {eq}x\sin \frac{1}{x} {/eq} if {eq}x\neq 0 {/eq} and by {eq}0 {/eq} if {eq}x = 0 {/eq}.

Let {eq}g(x) {/eq} be defined by {eq}x^2\sin \frac{1}{x} {/eq} if {eq}x\neq 0 {/eq} and by {eq}0 {/eq} if {eq}x = 0 {/eq}.

Show that {eq}f {/eq} is continuous, but not differentiable, at {eq}x = 0 {/eq}.

Show that {eq}g {/eq} is differentiable at {eq}0 {/eq}, and find {eq}{g}'(0) {/eq}.

Continuous Function:

Let {eq}f:A \to B {/eq} be a real valued function and {eq}c\in A {/eq}. Then the function {eq}f {/eq} is said to be continuous at the point {eq}x=c {/eq}, if for each {eq}\epsilon >0 {/eq} there exists a {eq}\delta >0 {/eq}, such that

{eq}\left| {f\left( x \right) - f\left( c \right)} \right| < \varepsilon {/eq} for all {eq}x {/eq} satisfying {eq}\left| {x - c} \right| < \delta {/eq}. That is {eq}\mathop {\lim }\limits_{x \to a} f\left( x \right) = f\left( c \right) {/eq}.

The function {eq}f {/eq} said to be differentiable at a point {eq}a\in A {/eq}, denoted by {eq}f'(a) {/eq}, if the limit {eq}\displaystyle \mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right) - f\left( a \right)}}{{x - a}}\left( {{\text{or by}}\,\,\mathop {\lim }\limits_{h \to 0} \frac{{f\left( {a + h} \right) - f\left( a \right)}}{h}} \right) {/eq} exists and {eq}\displaystyle \mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right) - f\left( a \right)}}{{x - a}} = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {a + h} \right) - f\left( a \right)}}{h} = f'\left( a \right) {/eq}

Answer and Explanation:

Let {eq}\epsilon >0 {/eq} be given.

We know that {eq}\left| {\sin x} \right| \leqslant 1,\,\,\,\forall \,\,x, {/eq} using this we have the following

{eq}\left| {f\left( x \right) - f(0)} \right| =\left| {f\left( x \right) - 0} \right| = \left| {x\sin \frac{1}{x}} \right| \leqslant \left| x \right| {/eq}.

So if we take {eq}\delta = \epsilon {/eq}, then {eq}\left| {f\left( x \right) - 0} \right| < \varepsilon {/eq}, for all {eq}x {/eq} satisfying {eq}\left| {x - 0} \right| = \left| x \right| < \delta {/eq}.

Thus we have for each {eq}\epsilon >0 {/eq} there exists a {eq}\delta {/eq} such that {eq}\left| {f\left( x \right) - 0} \right| < \varepsilon {/eq} for all x satisfying {eq}\left| x \right| < \delta {/eq}.

Hence we can say that the function {eq}f {/eq} is continuous at {eq}x=0 {/eq} and therefore {eq}\displaystyle \mathop {\lim }\limits_{x \to 0} f\left( x \right) = \mathop {\lim }\limits_{x \to 0} x\sin \frac{1}{x} = 0 = f\left( 0 \right) {/eq}.

Also we have the following

{eq}\displaystyle \mathop {\lim }\limits_{x \to 0} \frac{{f\left( x \right) - f\left( 0 \right)}}{{x - 0}} = \mathop {\lim }\limits_{x \to 0} \frac{{x\sin \frac{1}{x}}}{x} = \mathop {\lim }\limits_{x \to 0} \sin \frac{1}{x} {/eq}

If we aproach x=a using the seqence {eq}a_n=\frac{1}{n\pi} {/eq}, then we have {eq}\frac{1}{{n\pi }} \to 0,\,\,n \to \infty {/eq} and {eq}\sin a_n=\sin \frac{1}{{\frac{1}{{n\pi }}}} = \sin n\pi = 0,\,\,\,\forall \,\,n. {/eq} This indicates that {eq}\mathop {\lim }\limits_{n \to \infty }\sin a_n= \mathop {\lim }\limits_{n \to \infty } \sin \frac{1}{{\frac{1}{{n\pi }}}} = 0. {/eq}

But if we approach this point by using the sequence {eq}\displaystyle {b_n} = \frac{1}{{\left( {4n + 1} \right)\frac{\pi }{2}}} \to 0,\,\,{\text{as}}\,\,n \to \infty {/eq}, then we have {eq}\displaystyle \sin b_n=\sin \left( {4n + 1} \right)\frac{\pi }{2} = 1 {/eq}, for all {eq}n {/eq}, {eq}\displaystyle \mathop {\lim }\limits_{n \to \infty }\sin b_n=\mathop {\lim }\limits_{n \to \infty } \sin \left( {4n + 1} \right)\frac{\pi }{2} = 1 {/eq}.

So we can see that the limit is path dependent, which means the limit {eq}\displaystyle \mathop {\lim }\limits_{x \to 0} \sin \frac{1}{x} {/eq} does not exists.

Since limit {eq}\displaystyle \mathop {\lim }\limits_{x \to 0} \sin \frac{1}{x} {/eq} does not exists, we can say that {eq}f {/eq} is not differentiable at {eq}x=0 {/eq}.

To determine {eq}g'(0) {/eq}, we have the following

{eq}\displaystyle \mathop {\lim }\limits_{x \to 0} \frac{{g\left( x \right) - g\left( 0 \right)}}{{x - 0}} = \mathop {\lim }\limits_{x \to 0} \frac{{{x^2}\sin \frac{1}{x}}}{x} = \mathop {\lim }\limits_{x \to 0} x\sin \left( {\frac{1}{x}} \right) = 0,\,\,\,\left( {{\text{as}}\,\,{\text{we}}\,\,{\text{already}}\,\,{\text{prove }}\,{\text{that}}\,\,\mathop {\lim }\limits_{x \to 0} x\sin \left( {\frac{1}{x}} \right) = 0\,\,} \right) {/eq}

Hence we can say that {eq}g {/eq} is differentiable at {eq}x=0 {/eq} and {eq}g'(0)=0 {/eq}.


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