MATH 140 practice final exam with full solutions (McGill)

Direct substitution gives $\frac{\sqrt{4} - 2}{-1 + 1} = \frac{0}{0}$: the numerator and the denominator share the factor $x + 1$, and the root hides it on top. Two moves. Multiply top and bottom by the conjugate $\sqrt{x + 5} + 2$: the numerator becomes $(x + 5) - 4 = x + 1$. Factor the sum of cubes below: $x^3 + 1 = (x + 1)(x^2 - x + 1)$. For $x \ne -1$ the quotient is $\frac{x + 1}{(x + 1)(x^2 - x + 1)\left(\sqrt{x + 5} + 2\right)} = \frac{1}{(x^2 - x + 1)\left(\sqrt{x + 5} + 2\right)}$, continuous at $-1$, and direct substitution gives $\frac{1}{3 \cdot 4} = \frac{1}{12}$.

Substitution gives $\frac{\sin\pi}{0} = \frac{0}{0}$. The limit $\frac{\sin\theta}{\theta} \to 1$ needs an angle that tends to $0$, and $\pi x$ tends to $\pi$: writing $\frac{\sin(\pi x)}{\pi x} \to 1$ here is the classic error. Move: let $t = x - 1$, so $t \to 0$. Then $\sin(\pi x) = \sin(\pi + \pi t) = -\sin(\pi t)$ and $x^2 - 1 = (x - 1)(x + 1) = t(t + 2)$. The quotient is $-\frac{\sin(\pi t)}{\pi t} \cdot \frac{\pi}{t + 2} \to -1 \cdot \frac{\pi}{2} = -\frac{\pi}{2}$.

Move: take $x^2$ out of the root, remembering that $\sqrt{x^2} = |x|$: $\sqrt{16x^2 + 5} = |x|\sqrt{16 + \frac{5}{x^2}}$, then divide top and bottom by $x$. As $x \to \infty$, $|x| = x$: the quotient is $\frac{\sqrt{16 + 5/x^2} + 2}{3/x - 1} \to \frac{4 + 2}{-1} = -6$. As $x \to -\infty$, $|x| = -x$, so the numerator is $-x\sqrt{16 + \frac{5}{x^2}} + 2x = x\left(2 - \sqrt{16 + \frac{5}{x^2}}\right)$, and the quotient is $\frac{2 - \sqrt{16 + 5/x^2}}{3/x - 1} \to \frac{2 - 4}{-1} = 2$.

Move: divide top and bottom by $x$, the dominant power: $\frac{2 - \frac{\cos(x^2)}{x}}{1 + \frac{3}{x}}$. The quotient law needs the limit of $\frac{\cos(x^2)}{x}$, and $\cos(x^2)$ itself has none: it oscillates between $-1$ and $1$ forever. But it is bounded, and that is enough: for $x > 0$, $-\frac{1}{x} \le \frac{\cos(x^2)}{x} \le \frac{1}{x}$, and both bounds tend to $0$, so by the squeeze theorem $\frac{\cos(x^2)}{x} \to 0$. Now every piece has a limit and the quotient law applies: the limit is $\frac{2 - 0}{1 + 0} = 2$. Writing $\lim \cos(x^2)$ as if it existed, or dropping the cosine without the squeeze, costs the mark.

The first limit has the shape $\frac{g(a + h) - g(a)}{h}$ with $g(x) = \sqrt[3]{x}$ and $a = 27$, since the subtracted constant $3$ is exactly $g(27)$. So it equals $g'(27)$, and $g'(x) = \frac{1}{3}x^{-2/3}$ gives $\frac{1}{3} \cdot \frac{1}{9} = \frac{1}{27}$. The second has the shape $\frac{g(x) - g(a)}{x - a}$ with $g(x) = 2^x$, $a = 2$, and $g(2) = 4$ is indeed the subtracted constant. So it equals $g'(2) = 2^2\ln 2 = 4\ln 2$, since $(2^x)' = 2^x\ln 2$. Checking the subtracted constant is the whole test: $\lim_{x\to 2}\frac{2^x - 3}{x - 2}$ is not a derivative at all.

On $(-\infty, 4)$, $f$ is a polynomial; on $(4, \infty)$ it is $\sqrt{2x + 1}$ with $2x + 1 > 0$: continuous on both open pieces. Only the seam remains. $f(4) = \sqrt{9} = 3$, the right-hand limit is $3$, and the left-hand limit is $\lim_{x\to 4^-}(ax^2 + b) = 16a + b$. So $f$ is continuous on $\mathbb{R}$ if and only if $16a + b = 3$.

A differentiable function is continuous, so $16a + b = 3$ is required first. Right quotient, $h > 0$: $\frac{f(4 + h) - f(4)}{h} = \frac{\sqrt{9 + 2h} - 3}{h} = \frac{(9 + 2h) - 9}{h\left(\sqrt{9 + 2h} + 3\right)} = \frac{2}{\sqrt{9 + 2h} + 3} \to \frac{2}{6} = \frac{1}{3}$ (conjugate). Left quotient, $h < 0$: $f(4 + h) = a(16 + 8h + h^2) + b = 3 + 8ah + ah^2$ by the condition of b), so $\frac{f(4 + h) - f(4)}{h} = \frac{8ah + ah^2}{h} = 8a + ah \to 8a$.

With $a = 0$, $b = 3$: $16a + b = 3$, so $f$ is continuous at $4$. The left quotient is $8a = 0$ (the left piece is the constant $3$), the right one is $\frac{1}{3}$: two finite, different one-sided derivatives, so $f$ is not differentiable at $4$ and the graph has a CORNER at $(4, 3)$. If $16a + b \ne 3$, the numerator of the left quotient tends to $16a + b - 3 \ne 0$ while $h \to 0^-$: the quotient is infinite, and $f$ has no derivative, as it must, since it is not even continuous. Continuity is necessary for differentiability, never sufficient.

Form: $\frac{1 - 1 - 0}{0} = \frac{0}{0}$, indeterminate. Round 1, top and bottom differentiated separately: $\frac{e^x - \cos x}{2x}$, whose form is $\frac{1 - 1}{0} = \frac{0}{0}$ again. Round 2: $\frac{e^x + \sin x}{2}$, whose form is $\frac{1 + 0}{2}$: NOT indeterminate, so stop and substitute: the limit is $\frac{1}{2}$. A third round would differentiate the constant $2$ into $0$ and produce nonsense: the rule is applied only to an indeterminate form.

Form: $\infty \cdot 0$, since $\arctan x \to \frac{\pi}{2}$. Rewrite as a quotient, putting the factor $x$ in the denominator as $\frac{1}{x}$: $\frac{\frac{\pi}{2} - \arctan x}{1/x}$, of the form $\frac{0}{0}$. Differentiate: $\frac{-\frac{1}{1 + x^2}}{-\frac{1}{x^2}} = \frac{x^2}{1 + x^2} \to 1$. The limit is $1$: the gap $\frac{\pi}{2} - \arctan x$ shrinks exactly like $\frac{1}{x}$. The other rewriting, $\frac{x}{1/(\pi/2 - \arctan x)}$, leads to a heavier quotient.

As $x \to 1$ the base tends to $1$ and the exponent $\frac{1}{1 - x}$ to $+\infty$ from the left and $-\infty$ from the right: the form $1^{\pm\infty}$ is indeterminate, and the answer is NOT $1$. Near $1$, $x > 0$, so let $y = x^{1/(1 - x)}$ and $\ln y = \frac{\ln x}{1 - x}$, of the form $\frac{0}{0}$. Differentiate: $\frac{1/x}{-1} \to -1$, from both sides. Since the exponential is continuous, $y = e^{\ln y} \to e^{-1}$.

As $x \to 0^+$, $e^x - 1 \to 0^+$ and the exponent $x \to 0$: the form $0^0$. Let $y = (e^x - 1)^x$, so $\ln y = x\ln(e^x - 1)$, of the form $0 \cdot (-\infty)$. Rewrite: $\frac{\ln(e^x - 1)}{1/x}$, of the form $\frac{-\infty}{\infty}$. Differentiate, with the chain rule on the top: $\frac{e^x/(e^x - 1)}{-1/x^2} = -x e^x \cdot \frac{x}{e^x - 1}$. Now $\frac{x}{e^x - 1} \to 1$ (it is the reciprocal of $\frac{e^x - e^0}{x - 0} \to (e^x)'(0) = 1$), and $-x e^x \to 0$: so $\ln y \to 0$ and $y \to e^0 = 1$. Simplifying after the round avoids a second, heavier round.

Product rule with $u = x^4$ and $v = e^{-x/2}$. For $v'$, the chain rule with inner function $w = -\frac{x}{2}$: $v' = e^{w} \cdot w' = -\frac{1}{2}e^{-x/2}$. So $f'(x) = 4x^3 e^{-x/2} - \frac{1}{2}x^4 e^{-x/2} = \frac{1}{2}x^3 e^{-x/2}(8 - x)$. Horizontal tangents where $f'(x) = 0$: the exponential is never $0$, so $x^3 = 0$ or $8 - x = 0$, that is $x = 0$ or $x = 8$. The points are $(0, 0)$ and $(8, 8^4 e^{-4}) = (8, 4096e^{-4})$. The common slip is $v' = e^{-x/2}$, forgetting the factor $-\frac{1}{2}$ of the inner function: it moves the second point to $x = -4$.

Quotient rule with $N = \sin x - x\cos x$ and $D = \cos x + x\sin x$. Each needs the product rule on its second term: $N' = \cos x - (\cos x - x\sin x) = x\sin x$ and $D' = -\sin x + (\sin x + x\cos x) = x\cos x$. Then $g' = \frac{N'D - ND'}{D^2} = \frac{x\sin x(\cos x + x\sin x) - (\sin x - x\cos x)\,x\cos x}{D^2} = \frac{x\left[\sin x\cos x + x\sin^2 x - \sin x\cos x + x\cos^2 x\right]}{D^2} = \frac{x^2(\sin^2 x + \cos^2 x)}{D^2} = \frac{x^2}{(\cos x + x\sin x)^2}$, wherever $D \ne 0$.

Layers, from the outside: the square root $\sqrt{u}$, with $u = x + \sqrt{x}$, and inside $u$ the root $\sqrt{x}$. Chain rule on the outer root, then the sum rule inside: $h'(x) = \frac{1}{2\sqrt{x + \sqrt{x}}} \cdot \left(1 + \frac{1}{2\sqrt{x}}\right) = \frac{2\sqrt{x} + 1}{4\sqrt{x}\sqrt{x + \sqrt{x}}}$. At $x = 1$: $\frac{3}{4 \cdot 1 \cdot \sqrt{2}} = \frac{3}{4\sqrt{2}} = \frac{3\sqrt{2}}{8}$.

At $(2, 0)$: $2e^0 + 0 = 2$, so $P$ is on $C$. Differentiate both sides with respect to $x$, $y$ being a function of $x$. The term $x e^y$ is a product, and $e^y$ is a composition with inner function $y$: $\frac{d}{dx}(x e^y) = e^y + x e^y y'$. So $e^y + x e^y y' + y' = 0$, hence $y'(x e^y + 1) = -e^y$ and $\frac{dy}{dx} = -\frac{e^y}{x e^y + 1}$ wherever $x e^y + 1 \ne 0$. At $P$: $-\frac{1}{2 + 1} = -\frac{1}{3}$. Tangent: $y = -\frac{1}{3}(x - 2)$. The formula contains $y$: it gives the slope at a POINT, which is all that is needed.

A horizontal tangent needs the numerator $-e^y$ to vanish, which never happens: no horizontal tangent anywhere. A vertical tangent needs $x e^y + 1 = 0$ at a point of $C$, with the numerator nonzero. On $C$, $x e^y = 2 - y$, so the condition reads $2 - y + 1 = 0$: $y = 3$, then $x = (2 - 3)e^{-3} = -e^{-3}$. There the numerator is $-e^3 \ne 0$: $V = (-e^{-3}, 3)$ is the only point with a vertical tangent. It lies a hair to the left of the $y$-axis, $e^{-3}$ being about $0.05$, which is why the figure barely shows the curve turning back. Check: $C$ is the graph of $x = (2 - y)e^{-y}$, and $\frac{dx}{dy} = -(3 - y)e^{-y}$ vanishes exactly at $y = 3$.

Differentiate $e^y + x e^y y' + y' = 0$ once more, remembering that $y$ and $y'$ are both functions of $x$: $e^y y' + \left[e^y y' + x e^y (y')^2 + x e^y y''\right] + y'' = 0$, that is $2e^y y' + x e^y (y')^2 + (x e^y + 1)y'' = 0$. At $P$, with $y' = -\frac{1}{3}$: $-\frac{2}{3} + \frac{2}{9} + 3y'' = 0$, so $3y'' = \frac{4}{9}$ and $y'' = \frac{4}{27}$. It is positive: near $P$ the curve is concave up and lies ABOVE its tangent, as in the figure.

$x$ and $\arctan x$ are both increasing on $\mathbb{R}$, so their sum is increasing, and an increasing function never takes the same value twice: $f$ is one-to-one. No formula for $f^{-1}$ exists in terms of elementary functions, and none is needed. $f(0) = 0$, so $f^{-1}(0) = 0$; $f'(x) = 1 + \frac{1}{1 + x^2}$, the derivative of $\arctan$ being $\frac{1}{1 + x^2}$, and $f'(0) = 2 \ne 0$. Hence $(f^{-1})'(0) = \frac{1}{f'(f^{-1}(0))} = \frac{1}{2}$, and the tangent to $y = f^{-1}(x)$ at the origin is $y = \frac{x}{2}$, the reflection in $y = x$ of the tangent $y = 2x$ of $f$.

The power rule $(x^n)' = nx^{n-1}$ needs a CONSTANT exponent, and the exponential rule $(b^x)' = b^x\ln b$ a CONSTANT base; here both move. Take logarithms, $y > 0$: $\ln y = \sqrt{x}\ln x$. Differentiate both sides, the left by the chain rule, the right by the product rule: $\frac{y'}{y} = \frac{1}{2\sqrt{x}}\ln x + \sqrt{x} \cdot \frac{1}{x} = \frac{\ln x + 2}{2\sqrt{x}}$. So $y' = x^{\sqrt{x}}\,\frac{\ln x + 2}{2\sqrt{x}}$.

On $(-1, \infty)$ every factor is positive ($e^{2x}$, the root, and $1 + x$), so $y > 0$ and $\ln y$ can be taken without absolute values. Expand BEFORE differentiating: $\ln y = 2x + \frac{1}{2}\ln(1 + x^2) - 3\ln(1 + x)$. Differentiate: $\frac{y'}{y} = 2 + \frac{1}{2} \cdot \frac{2x}{1 + x^2} - \frac{3}{1 + x} = 2 + \frac{x}{1 + x^2} - \frac{3}{1 + x}$. The product and quotient rules would need three factors and a cube in the denominator; here each factor contributes one relative rate.