Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2018 / Paper 1 / 19

For statement (1), continuity at the interior point meansExplicitly, for every $\epsilon >0$ there exists $\delta >0$ such that $|x-x_0|<\delta$ implies $|f(x)-f(x_0)|<\epsilon$ for points in the domain. This is the epsilon-delta definition of continuity.

For statement (2), differentiability means the two-sided difference quotient has a finite limit:The same finite value must be obtained from positive and negative $h$. Differentiability implies continuity, whereas the converse need not hold.

For $m=0$, continuity fails. Take $x_k=(\pi /2+2\pi k{)}^{-1}$ and ${\tilde{x}}_k=(3\pi /2+2\pi k{)}^{-1}$. Both tend to zero, but the corresponding function values are $1$ and $-1$. Thus there is no limit, and the function is not differentiable either, since differentiability would imply continuity.

For $m=1$, the bound $|x\sin (1/x)|\le |x|$ and the squeeze theorem give continuity at zero. Its difference quotient is $\sin (1/h)$, however, which has no limit by the same two sequences. It is therefore not differentiable there.

For $m=2$, the bound $|x^2\sin (1/x)|\le x^2$ proves continuity. Its difference quotient is $h\sin (1/h)$, whose absolute value is bounded by $|h|$, so the derivative exists and equals zero. The complete classification is

For the positive terms $a_n=n!/2^n$, the ratio is $a_{n+1}/a_n=(n+1)/2\to \infty$. The ratio test therefore gives divergence. More directly, the terms increase by a factor at least two for all $n\ge 3$, so they fail the necessary term test for divergence condition $a_n\to 0$. Hence

Rationalize the positive difference:Since the denominator is at least $2n^2$,The upper bound is a constant multiple of the convergent p-series with exponent two. The comparison test therefore provesThe asymptotic equivalence of its summand to $a^2/(2n^2)$ gives the same conclusion by limit comparison test.