Corrected exercises: infinite limits, limits at infinity and asymptotes (MATH 140)

a) On the figure, the curve plunges down along $x = -2$ on the left and shoots up on the right; along $x = 1$ it climbs on BOTH sides; and it flattens onto the $x$-axis at both ends. So $\lim_{x\to -2^-} f(x) = -\infty$, $\lim_{x\to -2^+} f(x) = +\infty$, $\lim_{x\to 1} f(x) = +\infty$ and $\lim_{x\to\pm\infty} f(x) = 0$. From the formula, near $x = -2$: the numerator $x + 3 \to 1 > 0$ and $(x-1)^2 \to 9 > 0$, so the sign is the sign of $x + 2$, which tends to $0$ through negative values on the left ($0^-$) and through positive values on the right ($0^+$). Hence $\frac{1}{9 \cdot 0^-}$ gives $-\infty$ and $\frac{1}{9 \cdot 0^+}$ gives $+\infty$. Near $x = 1$: $x + 3 \to 4$, $x + 2 \to 3$, and $(x-1)^2 \to 0^+$ from BOTH sides, because a square is never negative: the limit is $+\infty$ on each side, so $\lim_{x\to 1} f(x) = +\infty$. The two lines $x = -2$ and $x = 1$ are vertical asymptotes. The factor to watch is the one that tends to $0$, and the power on it decides: odd power, the sign flips across the asymptote; even power, it does not.

b) As $x \to 3$, the numerator $2x \to 6 > 0$. For $x < 3$, $x - 3 < 0$, so the quotient is (positive)/(small negative): $\lim_{x\to 3^-} \frac{2x}{x-3} = -\infty$. For $x > 3$, $x - 3 > 0$: $\lim_{x\to 3^+} \frac{2x}{x-3} = +\infty$. The two one-sided limits are different, so $\lim_{x\to 3} \frac{2x}{x-3}$ does not exist, and we may not even write $= \infty$: that notation is reserved for a function that goes to $+\infty$ on both sides. Writing $\frac{6}{0} = \infty$ and stopping there is the most common way to lose this mark: the sign study IS the answer.

c) The numerator tends to $-1 - 5 = -6 < 0$. The denominator $(x+1)^2$ tends to $0$ and is positive on both sides of $-1$, so it tends to $0^+$. Then $\frac{x-5}{(x+1)^2}$ is (about $-6$)/(small positive) on both sides: $\lim_{x\to -1} \frac{x-5}{(x+1)^2} = -\infty$. A negative numerator over $0^+$ gives $-\infty$: the sign of the NUMERATOR matters as much as that of the denominator, and a student who only looks at the square answers $+\infty$.

d) Factor before deciding: $x^2 - 4 = (x-2)(x+2)$ and $x^2 - x - 2 = (x-2)(x+1)$. For $x \ne 2$, $h(x) = \frac{x+2}{x+1}$. At $x = 2$, both numerator and denominator of the original formula vanish: this is a form $\frac{0}{0}$, not a vertical asymptote. After simplification, $\lim_{x\to 2} h(x) = \frac{2+2}{2+1} = \frac{4}{3}$: the graph has a HOLE at $\left(2, \frac{4}{3}\right)$. At $x = -1$, the simplified numerator $x + 2 \to 1 \ne 0$ while $x + 1 \to 0$: on the left $x + 1 \to 0^-$ and $h(x) \to -\infty$; on the right $x + 1 \to 0^+$ and $h(x) \to +\infty$. The line $x = -1$ is a vertical asymptote. The rule: a zero of the denominator gives a vertical asymptote only when the numerator does NOT tend to $0$ there; when both tend to $0$, factor first.

e) As $x \to 2^+$, $x - 2 \to 0^+$ and $\ln u \to -\infty$ as $u \to 0^+$, so $\lim_{x\to 2^+} \ln(x-2) = -\infty$ (and the limit from the left makes no sense, $\ln(x-2)$ is not defined there). Next, $\tan x = \frac{\sin x}{\cos x}$ with $\sin x \to 1$ as $x \to \frac{\pi}{2}$; $\cos x > 0$ just left of $\frac{\pi}{2}$ and $< 0$ just right, so $\tan x \to +\infty$ on the left and $-\infty$ on the right. Finally, as $x \to 0^-$, $\frac{1}{x} \to -\infty$, and $e^u \to 0$ as $u \to -\infty$, so $e^{1/x} \to 0$; as $x \to 0^+$, $\frac{1}{x} \to +\infty$ and $e^{1/x} \to +\infty$. The vertical asymptotes are $x = 2$, $x = \frac{\pi}{2}$ (and every $\frac{\pi}{2} + k\pi$), and $x = 0$ for $e^{1/x}$, although that last one is infinite from ONE side only. One infinite one-sided limit is enough for an asymptote.

a) Divide by $x^2$, the highest power of the denominator: $f(x) = \frac{3 - \frac{5}{x} + \frac{1}{x^2}}{2 + \frac{7}{x^2}}$. As $x \to \pm\infty$, every fraction $\frac{c}{x^k}$ tends to $0$, so $f(x) \to \frac{3}{2}$ at both ends, and $y = \frac{3}{2}$ is a horizontal asymptote. The crossing: $f(x) = \frac{3}{2}$ means $2(3x^2 - 5x + 1) = 3(2x^2 + 7)$, that is $6x^2 - 10x + 2 = 6x^2 + 21$, so $-10x = 19$ and $x = -\frac{19}{10}$. No contradiction: an asymptote describes what the graph does as $x \to \pm\infty$, and says nothing about finite $x$. Here, for $x < -\frac{19}{10}$ the curve is above the line and comes down onto it from above as $x \to -\infty$, exactly as the figure shows.

b) First limit: the highest power of the denominator is $x^3$: $\frac{\frac{4}{x^2} - \frac{1}{x^3}}{1 + \frac{2}{x^3}} \to \frac{0 - 0}{1 + 0} = 0$. Degree of the numerator below degree of the denominator always gives $0$. Second limit: divide by $x^3$: $\frac{5 - \frac{1}{x^2}}{\frac{1}{x^3} - 2} \to \frac{5}{-2} = -\frac{5}{2}$, and this is the same as $x \to +\infty$: when the degrees are equal, the limit is the ratio of the leading coefficients, sign included. Writing $\frac{5}{2}$ because the $-2x^3$ was read as $2x^3$ costs the mark: keep each coefficient with its sign.

c) Divide by $x^2$: $\frac{2x^3 - x}{5 - x^2} = \frac{2x - \frac{1}{x}}{\frac{5}{x^2} - 1}$. The denominator tends to $-1$ at both ends. As $x \to +\infty$, the numerator $2x - \frac{1}{x} \to +\infty$, so the quotient tends to $-\infty$. As $x \to -\infty$, the numerator tends to $-\infty$, and dividing by a number close to $-1$ gives $+\infty$. No horizontal asymptote: the numerator's degree is larger. The shortcut, and the check, is to keep the leading terms: $\frac{2x^3}{-x^2} = -2x$, which tends to $-\infty$ at $+\infty$ and to $+\infty$ at $-\infty$. Answering $\infty$ for both ends is the classic mistake: the sign changes with the direction because the dominant term $-2x$ is ODD.

d) Only the leading term of each factor matters: $(2x - 1)^2(x + 3) = x^3\left(2 - \frac{1}{x}\right)^2\left(1 + \frac{3}{x}\right)$ and $x^3 - 4 = x^3\left(1 - \frac{4}{x^3}\right)$. Cancel $x^3$: $\frac{\left(2 - \frac{1}{x}\right)^2\left(1 + \frac{3}{x}\right)}{1 - \frac{4}{x^3}} \to \frac{2^2 \cdot 1}{1} = 4$. Factoring $x$ out of each bracket is quicker and safer than expanding, and it shows at a glance that the numerator has degree $3$ with leading coefficient $2^2 \cdot 1 = 4$. The trap is to read the leading coefficient of $(2x-1)^2$ as $2$ instead of $4$.

e) As $x \to +\infty$, $x^3 \to +\infty$ and $-100x^2 \to -\infty$: this is the form $\infty - \infty$, which decides nothing by itself. Factor out the dominant power: $x^3 - 100x^2 = x^3\left(1 - \frac{100}{x}\right)$, a product of a factor tending to $+\infty$ and a factor tending to $1$, so the limit is $+\infty$. The first values are misleading: at $x = 50$ the polynomial equals $125000 - 250000 < 0$, and it only becomes positive for $x > 100$. As $x \to -\infty$, $x^3 \to -\infty$ AND $-100x^2 \to -\infty$: two terms going to $-\infty$ add up to $-\infty$, and there is nothing to decide. A form $\infty - \infty$ appears only when the two terms pull in opposite directions.

a) Divide the numerator and the denominator by $x$. For $x > 0$, $x = \sqrt{x^2}$, so $\frac{\sqrt{4x^2 + 1}}{x} = \sqrt{\frac{4x^2 + 1}{x^2}} = \sqrt{4 + \frac{1}{x^2}}$. Then $f(x) = \frac{6 - \frac{1}{x}}{\sqrt{4 + \frac{1}{x^2}}} \to \frac{6}{\sqrt 4} = 3$. The limit $\sqrt{4 + \frac{1}{x^2}} \to 2$ uses the continuity of the square root at $4$.

b) For $x < 0$, $\sqrt{x^2} = |x| = -x$, so $x = -\sqrt{x^2}$ and $\frac{\sqrt{4x^2 + 1}}{x} = -\sqrt{4 + \frac{1}{x^2}}$. Hence $f(x) = \frac{6 - \frac{1}{x}}{-\sqrt{4 + \frac{1}{x^2}}} \to -\frac{6}{2} = -3$. The graph has TWO horizontal asymptotes, $y = 3$ on the right and $y = -3$ on the left, which is what the figure shows. Sanity check before any algebra: for $x$ very negative, the numerator $6x - 1$ is negative and the root is positive, so $f(x) < 0$, and an answer of $+3$ is impossible. Forgetting the minus sign is the trap of this chapter, and it costs the whole limit.

c) This is the form $\infty - \infty$. Multiply and divide by the conjugate: $\sqrt{x^2 + 6x} - x = \frac{(x^2 + 6x) - x^2}{\sqrt{x^2 + 6x} + x} = \frac{6x}{\sqrt{x^2 + 6x} + x}$. Divide by $x > 0$: $\frac{6}{\sqrt{1 + \frac{6}{x}} + 1} \to \frac{6}{1 + 1} = 3$. Check with $x = 100$: $\sqrt{10600} \approx 102.96$ (since $103^2 = 10609$), and $102.96 - 100 \approx 2.96$, close to $3$.

d) As $x \to -\infty$, $\sqrt{x^2 + 6x} \to +\infty$ (the function is defined for $x \le -6$) and $-x \to +\infty$: the sum of two terms going to $+\infty$ goes to $+\infty$. There is no indeterminate form, and the conjugate would only complicate things. For the second limit, $\sqrt{x^2 + 6x} \to +\infty$ and $x \to -\infty$: now it IS a form $\infty - \infty$. Conjugate: $\sqrt{x^2 + 6x} + x = \frac{(x^2 + 6x) - x^2}{\sqrt{x^2 + 6x} - x} = \frac{6x}{\sqrt{x^2 + 6x} - x}$. Divide by $x < 0$, using $\frac{\sqrt{x^2 + 6x}}{x} = -\sqrt{1 + \frac{6}{x}}$: $\frac{6}{-\sqrt{1 + \frac{6}{x}} - 1} \to \frac{6}{-2} = -3$. Always decide first whether the form is indeterminate, then whether $x$ is positive or negative before bringing it under the root.

e) $\sqrt{4x^2 + x} - 2x = \frac{(4x^2 + x) - 4x^2}{\sqrt{4x^2 + x} + 2x} = \frac{x}{\sqrt{4x^2 + x} + 2x} = \frac{1}{\sqrt{4 + \frac{1}{x}} + 2} \to \frac{1}{2 + 2} = \frac{1}{4}$. The student's approximation is true as a RATIO, $\frac{\sqrt{4x^2 + x}}{2x} \to 1$, but a difference of two huge numbers is decided by what the approximation threw away: here $\sqrt{4x^2 + x}$ exceeds $2x$ by about $\frac{1}{4}$, a gap that does not shrink. Replacing a term by an equivalent one is legitimate in a product or a quotient, never in a difference.

a) As $x \to \infty$, $e^x$ dominates: divide by it. $g(x) = \frac{2 + 3e^{-x}}{1 - e^{-x}} \to \frac{2 + 0}{1 - 0} = 2$. As $x \to -\infty$, $e^x \to 0$ and nothing needs dividing: $g(x) \to \frac{0 + 3}{0 - 1} = -3$. Two different horizontal asymptotes, $y = 2$ on the right and $y = -3$ on the left, as on the figure. Dividing by $e^x$ at $-\infty$ would be dividing by a number that tends to $0$: the dominant term is not the same at both ends, and $e^x$ only dominates as $x \to +\infty$.

b) As $x \to 0$, the numerator tends to $2 + 3 = 5 > 0$ and the denominator $e^x - 1$ tends to $0$. Its sign: $e^x > 1$ for $x > 0$ and $e^x < 1$ for $x < 0$, so $e^x - 1 \to 0^+$ on the right and $0^-$ on the left. Hence $\lim_{x\to 0^-} g(x) = -\infty$ and $\lim_{x\to 0^+} g(x) = +\infty$. The asymptotes: $y = 2$ as $x \to \infty$, $y = -3$ as $x \to -\infty$, and $x = 0$, the $y$-axis itself. The sign of $e^x - 1$ comes from the fact that $e^x$ is increasing with $e^0 = 1$: say it, it is the whole justification.