Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2011 / Paper 1 / 20

Set $a=\pi /2$ for this part. In the interval $(-\pi ,\pi )$ the special points are $-a,0,a$. Elsewhere $f_0=\sin (\tan x)$ is a composition of smooth functions.

At zero, $\tan x=x+O(x^3)$ and therefore $\sin (\tan x)=x+O(x^3)$. Its assigned value is zero, so the function is continuous and differentiable there, with $f_0^{\prime }(0)=1$.

At either tangent pole $\pm a$, the bounded sine oscillates without a limit. For an explicit check at $a$, let $x_j=\arctan (2\pi j+\pi /2)$ and $u_j=\arctan (2\pi j+3\pi /2)$. Both approach $a$ from below, but the function values are respectively $1$ and $-1$. Negating these sequences gives the same obstruction at $-a$. Assigning the value zero at the poles cannot repair either limit.

Thus $f_0$ is discontinuous and nondifferentiable exactly at $x=\pm \pi /2$, and is smooth elsewhere in the specified interval.

At $-a$, the prefactor $(x-a)/x$ tends to $2$, so the nonvanishing amplitude leaves the oscillatory discontinuity intact. At zero, use $\sin (\tan x)=x+O(x^3)$ to obtainThe limit is $-a$, whereas the assigned value is zero. This is a removable discontinuity, not a continuous crossing.

At $a$, write $x=a+h$. The estimateproves continuity at the assigned zero. But its difference quotient isThe sequences from part (i) make this approach $1/a$ and $-1/a$, so the derivative does not exist. This illustrates the distinct continuity and differentiability thresholds for a damped oscillation.

Therefore the discontinuities are $x=-\pi /2$ and $x=0$; the nondifferentiability points are $x=-\pi /2,0,\pi /2$. The function is smooth elsewhere.

At $-a$, the squared prefactor tends to $4$, so the oscillatory discontinuity again remains. Near zero,It is unbounded near zero, despite its assigned zero value there, so it is neither continuous nor differentiable at zero.

At $a$, put $x=a+h$. ThenThus $f_2$ is continuous and differentiable at $x=\pi /2$, with $f_2^{\prime }(\pi /2)=0$. After translating the point to zero, the estimate shows how quadratic damping restores differentiability at the origin. The only discontinuities and nondifferentiability points are $x=-\pi /2$ and $x=0$; the function is smooth away from these two points and $\pi /2$. At $\pi /2$, its derivative exists but is not continuous.

Indeed, for $x\ne a$ near $a$, put $B(x)=(x-a{)}^2/x^2$. Then $f_2^{\prime }=B^{\prime }\sin (\tan x)+B{\sec }^{2}x\cos (\tan x)$. Here $B^{\prime }\to 0$ and $B{\sec }^{2}x\to 1/a^2$, so the derivative retains oscillations. Existence of the derivative at a point does not require continuity of nearby derivatives.

Let $u=x^{-5}$, which tends to zero as $|x|\to \infty$. Factor the denominator and use the cosecant expansion near zero:Substitution gives the first two nonzero terms and the requested omitted order:More precisely the next term is $7/(360x^{18})$, followed by $O(x^{-28})$. The outside factor $x^2$ is essential when converting the remainder order from powers of $u$ to powers of $x$.