Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2008 / Paper 1 / 19

Continuity at $x_0$ requires that $f(x_0)$ be defined and that ${\lim }_{x\to x_0}f(x)=f(x_0)$. In the epsilon-delta definition, for each $\epsilon >0$ there must be a $\delta >0$ such that $|f(x)-f(x_0)|<\epsilon$ whenever $|x-x_0|<\delta$ within the domain. Differentiability requires the finite two-sided limitIn particular, merely having a punctured limit of a function does not imply continuity unless the assigned value agrees with that limit.

For every $n\ge 1$, $|x^n\sin (1/x)|\le |x{|}^n\to 0$, so the squeeze theorem gives limit zero. For $n=0$, sequences $x_k=(\pi /2+2k\pi {)}^{-1}$ and ${\tilde{x}}_k=(3\pi /2+2k\pi {)}^{-1}$ give function values $1$ and $-1$, so no limit exists. HenceThe value assigned at zero when $n=4$ does not affect this punctured limit.

For $n=1,2,3$ the limit zero equals the assigned value at zero. For $n=4$ the limit is zero but the assigned value is one, giving a removable discontinuity. For $n=0$ there is no limit. Thus continuity at zero holds precisely for

For $n=1,2,3$, the difference quotient at zero is $x^{n-1}\sin (1/x)$. It oscillates without a limit for $n=1$, while the squeeze theorem gives limit zero for $n=2,3$. The other two cases fail continuity and therefore cannot be differentiable. Hence

If $A>B$, divide numerator and denominator by $e^{A/x^2}$. The expression becomes $(A+B{e}^{-(A-B)/x^2})/(1+e^{-(A-B)/x^2})$, which tends to $A$ since $A-B>0$. If $B>A$, the same argument gives $B$. If $A=B$, the expression equals $A$ for every $x\ne 0$. ThusThis is the two-term case of exponential concentration of a weighted mean, and holds for negative as well as positive parameters.

For a series with eventually nonzero terms $a_n$, suppose $r={\lim }_{n\to \infty }|a_{n+1}/a_n|$ exists. The ratio test says that $r<1$ implies absolute convergence, while $r>1$ implies divergence because the terms fail to tend to zero. At $r=1$ the test is inconclusive.

Here $|a_{n+1}/a_n|=n/(n+1)\to 1$. Therefore the ratio test is inconclusive. The series does in fact converge by the alternating series test, and fails to converge absolutely because the harmonic series diverges; these facts use other tests.

For the positive summand $a_n=(n!{)}^3{e}^{3n}/(3n)!$, the ratio test givesThus the series converges absolutely.