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  • The function f defined by <img alt="\mathrm{f(x)=\left\{\begin{array}{c} \frac{\sin x^2}{x}, x \neq 0 \\ 0, x=0 \end{array}\right. \text { is } }" src="https://entrancecorner.oncodecogs

The function f defined by
\mathrm{f(x)=\left\{\begin{array}{c} \frac{\sin x^2}{x}, x \neq 0 \\ 0, x=0 \end{array}\right. \text { is } }

Option: 1

continuous and derivable at x=0


Option: 2

neither continuous nor derivable at x=0
 


Option: 3

continuous but not derivable at x=0
 


Option: 4

none of these.


Answers (1)

best_answer

We have :
\mathrm{\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{\sin x^2}{x}=\lim _{x \rightarrow 0}\left(\frac{\sin x^2}{x^2}\right) x=1 \times 0=0=f(0) }
So, f(x) is continuous at x=0 . f(x) is also derivable at x=0, because
\mathrm{ \lim _{x \rightarrow 0} \frac{f(x)-f(0)}{x-0}=\lim _{x \rightarrow 0} \frac{\sin x^2}{x}=\lim _{x \rightarrow 0} \frac{\sin x^2}{x^2}=1 \text { exists finitely. } }

Posted by

avinash.dongre

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