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A function $\mathrm{f}$ is defined as follows
$\mathrm{f}(\mathrm{x})=\mathrm{x}^{\mathrm{p}} \cos \left(\frac{1}{\mathrm{x}}\right), \mathrm{x} \neq 0$
$f(0)=0$
What conditions should be imposed on p so that $\mathrm{f}$ may be continuous at $x=0$ ?
Options:
$\mathrm{f}(\mathrm{x})=\mathrm{x}^{\mathrm{p}} \cos \left(\frac{1}{\mathrm{x}}\right), \mathrm{x} \neq 0$
$f(0)=0$
What conditions should be imposed on p so that $\mathrm{f}$ may be continuous at $x=0$ ?
Solution:
1157 Upvotes
Verified Answer
The correct answer is:
$\mathrm{p}>0$
Given function is defined as:
$f(x)=\left\{\begin{array}{ll}x^{p} \cos \left(\frac{1}{x}\right) & x \neq 0 \\ 0, & x=0\end{array}\right.$
For continuity:
LHS $: \lim _{x \rightarrow 0} f(x)=$ RHS $\lim _{x \rightarrow 0} f(x)=f(0)$
$\Rightarrow \lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} x^{p} \cos \left(\frac{1}{x}\right)=0$
$\Rightarrow \lim _{x \rightarrow 0} x^{p} \cos \left(\frac{1}{x}\right)=0$
$\cos \left(\frac{1}{\mathrm{x}}\right)$ is always a finite quantity if $\mathrm{x} \rightarrow 0$
$\Rightarrow \mathrm{x}^{\mathrm{p}}=0$
which is possible only if $\mathrm{p}>0$.
$f(x)=\left\{\begin{array}{ll}x^{p} \cos \left(\frac{1}{x}\right) & x \neq 0 \\ 0, & x=0\end{array}\right.$
For continuity:
LHS $: \lim _{x \rightarrow 0} f(x)=$ RHS $\lim _{x \rightarrow 0} f(x)=f(0)$
$\Rightarrow \lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} x^{p} \cos \left(\frac{1}{x}\right)=0$
$\Rightarrow \lim _{x \rightarrow 0} x^{p} \cos \left(\frac{1}{x}\right)=0$
$\cos \left(\frac{1}{\mathrm{x}}\right)$ is always a finite quantity if $\mathrm{x} \rightarrow 0$
$\Rightarrow \mathrm{x}^{\mathrm{p}}=0$
which is possible only if $\mathrm{p}>0$.
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