Limits and Continuity: math test solutions, College – download the PDF

1 Computing limits / 4 pts

(a) \(\frac{(x-5)(x+5)}{x-5}=x+5\to10\). (1 pt)

(b) Conjugate: \(\dfrac{x}{x(\sqrt{x+4}+2)}=\dfrac{1}{\sqrt{x+4}+2}\to\dfrac14\). (1 pt)

(c) Divide by \(x^2\): \(\dfrac{7-1/x}{2+9/x^2}\to\dfrac72\). (1 pt)

(d) \(x^2+x-2=(x+2)(x-1)\), so the quotient is \(x+2\to3\). (1 pt)

2 A jump / 3 pts

(a) Left: \(2\cdot3-1=5\). Right: \(10-3=7\). (1 pt each, 2 pts)

(b) The one-sided limits differ, so the limit does not exist. (0.5 pt)

(c) \(g(3)=7\). The one-sided limits exist and differ: a jump discontinuity. (0.5 pt)

3 Asymptotes and a hole / 3 pts

Factor: \(r(x)=\dfrac{2(x+3)}{(x-3)(x+3)}=\dfrac{2}{x-3}\) for \(x\neq-3\).

(a) At \(x=3\) the denominator is 0 and the numerator is 12, so \(x=3\) is a vertical asymptote. (1 pt)

(b) At \(x=-3\) the common factor cancels: the limit is \(\dfrac{2}{-6}=-\dfrac13\), so there is a hole at \(\left(-3,-\frac13\right)\). (1 pt)

(c) \(r(x)\to0\) as \(x\to\pm\infty\), so \(y=0\) is the horizontal asymptote. (1 pt)

4 Squeeze / 3 pts

Since \(-1\le\sin x\le1\), \(3\le4+\sin x\le5\); dividing by \(x^2>0\) gives the inequality. (1 pt)

Both \(\frac3{x^2}\) and \(\frac5{x^2}\) tend to 0 as \(x\to+\infty\). (1 pt)

By the squeeze theorem, the limit is 0. (1 pt)

5 Continuity at a point / 4 pts

(a) \(x^2-x-6=(x-3)(x+2)\), so \(f(x)=x+2\) for \(x\neq3\) and the limit is 5. (2 pts)

(b) The limit exists (5) but differs from \(f(3)=1\): a removable discontinuity. (1 pt)

(c) Continuity requires \(k=5\). (1 pt)

6 Intermediate Value Theorem / 3 pts

(a) \(h\) is a polynomial, hence continuous. \(h(1)=1+2-5=-2<0\) and \(h(2)=8+4-5=7>0\). By the IVT, there is a root in \((1,2)\). (2 pts)

(b) \(h(1)=-2<0\) and \(h(1.5)=1.375>0\), so a root lies in \((1,1.5)\). (1 pt)