Science:Math Exam Resources/Courses/MATH100/December 2015/Question 07

MATH100 December 2015

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Question 07
Is the function $f(x)={\begin{cases}{\sqrt {1+x^{2}}}-1&{\text{if }}x\leq 0,\\x^{2}\cos {\left({\frac {1}{x}}\right)}&{\text{if }}x>0\end{cases}}$ differentiable at $x=0$? You must explain your answer using the definition of the derivative.

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Hint
Recall the definition of differentiability at some point.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. First, we check that $f$ is continuous at 0. (Note that if $f$ is not continuous at 0, $f$ is not differentiable at 0.) Using $\left|\cos {\left({\frac {1}{x}}\right)}\right|\leq 1$, we can compute the right hand limit as $0\leq \lim _{x\to 0+}\left|x^{2}\cos {\left({\frac {1}{x}}\right)}\right|\leq \lim _{x\to 0+}|x^{2}|=0\implies \lim _{x\to 0+}f(x)=\lim _{x\to 0+}x^{2}\cos {\left({\frac {1}{x}}\right)}=0.$ On other hand, it is easy to get $\lim _{x\to 0-}f(x)=\lim _{x\to 0-}{\sqrt {1+x^{2}}}-1={\sqrt {1+0}}-1=0=f(0).$ Therefore, $f$ is continuous at 0. Now, we show $f$ is differentiable at 0. By the definition of derivative, first consider the right hand limit: $\lim _{x\to 0+}{\frac {f(x)-f(0)}{x-0}}=\lim _{x\to 0+}{\frac {x^{2}\cos {\left({\frac {1}{x}}\right)}}{x}}=\lim _{x\to 0+}x\cos {\left({\frac {1}{x}}\right)}=0.$ The last inequality follows from $0\leq \lim _{x\to 0+}|x\cos {\left({\frac {1}{x}}\right)}|\leq \lim _{x\to 0+}|x|=0.$ Then, the left hand limit can be computed as $\lim _{x\to 0-}{\frac {f(x)-f(0)}{x-0}}=\lim _{x\to 0-}{\frac {{\sqrt {1+x^{2}}}-1}{x}}=\lim _{x\to 0-}{\frac {{\sqrt {1+x^{2}}}-1}{x}}\cdot {\frac {{\sqrt {1+x^{2}}}+1}{{\sqrt {1+x^{2}}}+1}}=\lim _{x\to 0-}{\frac {({\sqrt {1+x^{2}}})^{2}-1}{x({\sqrt {1+x^{2}}}+1)}}=\lim _{x\to 0-}{\frac {x}{{\sqrt {1+x^{2}}}+1}}=0.$ Since the right and left hand limits coincide, the limit exists so that $f$ is differentiable at 0 and $\color {blue}{f'(0)=0}$.