Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics II / 6

The function $f$ is differentiable at $c$ if the finite limit of a functionexists. If it exists, put $L=f^{\prime }(c)$ and defineThe definition of derivative gives $ϵ(x)\to 0$ as $x\to c$. For $x\ne c$, rearrangement gives the desired local linear approximationAt $x=c$, this identity holds because both terms containing $x-c$ vanish. Conversely, if such an identity holds, then for $x\ne c$ its difference quotient is $L+ϵ(x)\to L$. Hence $f$ is differentiable at $c$, and necessarily $\boxed{L=f^{\prime }(c)}$.

Use the local linear approximation from part (a), with $L=f^{\prime }(c)$. Since $x_n\le c\le y_n$ and $x_n\ne y_n$, we have $y_n-x_n>0$, and subtraction givesThe coefficients $(y_n-c)/(y_n-x_n)$ and $(c-x_n)/(y_n-x_n)$ are nonnegative and sum to one. ThereforeHere $ϵ(c)=0$ handles any endpoint of a secant that happens to equal $c$. Thus secants straddling a differentiability point satisfy

No. On $(-1,1)$ take $c=0$ andAt zero, its difference quotient has absolute value $|x\sin (1/x)|\le |x|\to 0$, so $f^{\prime }(0)=0$. Away from zero, the product rule and chain rule apply. DefineThese are distinct positive inputs tending to zero. Since $\sin A_n=0$ and $\sin (A_n+B_n)=\sin B_n$,The even-indexed subsequence tends to $-2/\pi$, whereas the odd-indexed subsequence tends to $2/(3\pi )$. Thus the limit of a sequence does not exist. The two points lie on the same side of $c$, so the nonnegative-weight estimate in part (b) is unavailable. This is a concrete instance of differentiability does not imply continuity of the derivative.