Question

A function f is defined as follows : $f\left(x\right)=x^p\cos \left(1/x\right),x\ne 0,f\left(0\right)=0.$ What conditions should be imposed on $p$ so that (i) $f$ may be continuous at $x=0$, (ii) $f$ may have a differential coefficient at $x=0$?

• A. $p$ can be any real number
• B. $p$ should be greater than 1
• C. $p$ should be less than 1
• D. $p$ can be any real number except 1

Answer 1

Answer: (B)

$p$ should be greater than 1

$f(0+0)=\underset{h\to 0}{lim}{(0+h)}^p\cos \frac{1}{0+h}=\underset{h\to 0}{lim}{(-h)}^p\cos \frac{1}{h}...\left(1\right)$

And $f(0+0)=\underset{h\to 0}{lim}{(-h)}^p\cos \left(\frac{1}{-h}\right)=\underset{h\to 0}{lim}{(-h)}^p\cos \frac{1}{h}...\left(2\right)$

Now in order that the function may be continuous at x=0,the limits given in (1) and (2) must both tend to zero.
This will be the case if p>0 which is the required condition.
Again, $R{f}^{\prime }\left(0\right)=\underset{h\to 0}{lim}\frac{h^p\cos \left(1/h\right)-0}{h}=\underset{h\to 0}{lim}{h}^{p-1}\cos \frac{1}{h}...\left(3\right)$

And $L{f}^{\prime }\left(0\right)=\underset{h\to 0}{lim}\frac{{(-h)}^p\cos (-1/h)-0}{-h}=\underset{h\to 0}{lim}-{(-1)}^p{h}^{p-1}\cos \left(1/h\right)....\left(4\right)$

Now in order that f'(0) may exist,it is necessary that the limits in (3) and (4) must tend to the same quantity.
This will be the case when p>1 for in that case both $R{f}^{\prime }\left(0\right)andL{f}^{\prime }\left(0\right)$ will be zero.

Hence in order that f may have a differential coefficient at $x=0$, p should be greater than 1.