Chapter 2 Solutions

Authored worked explanation for the original source problem.

1. At a general point a, $f\left(x\right)-f\left(a\right)=4(x-a)$.

   Why: The constant 1 cancels in the change of output.

2. Thus $\left|f\right(x)-f(a\left)\right|=4|x-a|<4\delta$ when $|x-a|<\delta$.

   Why: Multiplication by 4 scales every input error by 4.

Reviewed answer: $\epsilon =4\delta$; to guarantee a prescribed ε use $\delta =\epsilon /4$.

Takeaway: For a linear function, error scales by the absolute slope.

Authored worked explanation for the original source problem.

1. Rewrite the expression as $(4-3/x^2)/(5+2/x+5/x^2)$.

   Why: Divide each polynomial by its highest power before taking the outer quotient.

2. The reciprocal powers tend to 0, leaving 4/5.

   Why: The limiting denominator is 5, so the quotient limit rule is valid.

Reviewed answer: $4/5$

Takeaway: Leading coefficients determine this rational limit.

Authored worked explanation for the original source problem.

1. For nearby inputs other than 4, the function is $\sqrt{x}$.

   Why: A limit ignores an isolated value at the target input.

2. So ${\lim }_{x\to 4}f\left(x\right)=\sqrt{4}=2$, even though $f\left(4\right)=4$.

   Why: The nearby branch is continuous but the assigned point creates a discontinuity.

Reviewed answer: $2$

Takeaway: The value at one point and the limit there can disagree.

Authored worked explanation for the original source problem.

1. Rationalize with $\cos x+1$: $(\cos x-1)/\sin x=-\sin x/(\cos x+1)$ near zero.

   Why: Use cos²−1=−sin² and cancel only where sine is nonzero.

2. The numerator tends to 0 and denominator to 2, giving 0.

   Why: The rewritten denominator is nonzero at the limit point.

Reviewed answer: $0$

Takeaway: Trig identities can remove an apparent 0/0 without derivatives.

Authored worked explanation for the original source problem.

1. Use $1-{\cos }^{2}x={\sin }^{2}x$ and $\sin \left(2x\right)=2\sin x\cos x$.

   Why: These identities factor the numerator and denominator.

2. The quotient becomes $\sin x/(1+\cos x)$ near zero, which tends to 0/2=0.

   Why: Cancelling the common sine factor is valid in the punctured neighborhood.

Reviewed answer: $0$

Takeaway: Factor first; evaluate the equivalent nearby expression second.

Authored worked explanation for the original source problem.

Source review note: The source gives 1/2. The printed numerator contains +sin x, so it leaves a 1/x divergence and has no two-sided limit.

1. Split the quotient into $\cos x(\cos x-1)/x^2+\sin x/x^2$.

   Why: The two terms have different sizes near zero.

2. The first term tends to −1/2 because $\cos x-1=-2{\sin }^2(x/2)$ and $\sin u/u\to 1$.

   Why: This uses special trig limits already introduced in this chapter.

3. The second term is $(\sin x/x)/x$, which behaves like 1/x.

   Why: Its sign changes across zero and its magnitude grows without bound.

4. A bounded first term cannot cancel that divergence; the one-sided limits are opposite infinities.

   Why: A finite two-sided limit requires agreement from both sides.

Reviewed answer: No two-sided limit; right limit $+\infty$, left limit $-\infty$.

Takeaway: Check the lowest-order behavior of every numerator term.

Authored worked explanation for the original source problem.

Source review note: The source gives +1/2; rationalization of the printed 4−x radicand yields −1/2.

1. $\sqrt{4-x}-2=-x/(\sqrt{4-x}+2)$ and $\sqrt{1+x}-1=x/(\sqrt{1+x}+1)$.

   Why: Multiply each difference by its conjugate.

2. Cancel x for x≠0 to get $-(\sqrt{1+x}+1)/(\sqrt{4-x}+2)$.

   Why: The negative sign from the first radicand remains.

3. At zero this tends to −2/4=−1/2.

   Why: The rewritten denominator does not vanish.

Reviewed answer: $-1/2$

Takeaway: A conjugate can expose a sign as well as a common factor.

Authored worked explanation for the original source problem.

1. Rationalize both differences: the numerator is $(x^2-1)/(\sqrt{x^2+3}+2)$ and denominator is $(x^2-1)/(\sqrt{x^2+8}+3)$.

   Why: Both have the same vanishing polynomial factor.

2. Cancel it near x=1 to obtain $(\sqrt{x^2+8}+3)/(\sqrt{x^2+3}+2)$.

   Why: A punctured neighborhood excludes x=1 but allows evaluating the equivalent expression's limit.

3. Substitute 1: 6/4=3/2.

   Why: The new denominator stays positive.

Reviewed answer: $3/2$

Takeaway: Rationalizing both numerator and denominator can reveal the same factor.

Authored worked explanation for the original source problem.

Source review note: The source gives 4/3, but its printed denominator has x²+14, which does not vanish at x=2. The reviewed limit is 0.

1. At x=2 the numerator is $\sqrt{9}-3=0$, but the denominator is $\sqrt{18}-4=3\sqrt{2}-4\ne 0$.

   Why: Check the expression as printed before assuming it is 0/0.

2. Continuity of the radicals and nonzero denominator give the limit 0.

   Why: The quotient rule for limits applies directly.

Reviewed answer: $0$

Takeaway: Not every radical difference needs rationalization.

Authored worked explanation for the original source problem.

1. For $f\left(x\right)=c$, $\left[f\right(a+h)-f(a\left)\right]/h=(c-c)/h=0$ for every h≠0.

   Why: The derivative uses a difference quotient before taking the limit.

2. Its limit as h approaches zero is 0.

   Why: We never divide by h=0; the nearby values are already identically zero.

Reviewed answer: $0$

Takeaway: A derivative is a limit of quotients, not a quotient evaluated at zero.

Authored solution to an added practice problem.

1. A polynomial is continuous at 2, so substitute directly.

   Why: No denominator or domain obstruction occurs.

2. $2^2+2·2-1=7$.

   Why: Evaluate after confirming continuity.

Reviewed answer: $7$

Takeaway: Direct substitution is the first tool to try.

Authored solution to an added practice problem.

1. Factor $x^2-9=(x-3)(x+3)$.

   Why: The numerator and denominator share a vanishing factor.

2. For x≠3, the quotient is x+3, so its limit is 6.

   Why: Cancelling changes the point value but preserves nearby values.

Reviewed answer: $6$

Takeaway: A hole need not prevent a limit.

Authored solution to an added practice problem.

1. Multiply by $\sqrt{9+x}+3$ to get $1/(\sqrt{9+x}+3)$ for x≠0.

   Why: The conjugate changes the numerator into x.

2. The rewritten expression tends to 1/6.

   Why: Its denominator is continuous and nonzero at zero.

Reviewed answer: $1/6$

Takeaway: Conjugates turn a radical difference into a cancelable factor.

Authored solution to an added practice problem.

1. For x<1, the branch is x+2, giving a left limit of 3.

   Why: Use the formula valid on that side.

2. For x>1, the branch is 2x+3, giving a right limit of 5.

   Why: A separate branch controls the other side.

3. Since 3≠5, no two-sided limit exists, whatever value is assigned at x=1.

   Why: Changing one point cannot join two different nearby heights.

Reviewed answer: Left limit 3; right limit 5; two-sided limit does not exist.

Takeaway: Compare sides before claiming a two-sided limit.

Authored solution to an added practice problem.

1. For x≠2, $(x^2-4)/(x-2)=x+2$.

   Why: Factor and cancel away from the missing point.

2. Its limit at 2 is 4, so define f(2)=4.

   Why: Continuity requires the point value to equal the existing limit.

Reviewed answer: $k=4$.

Takeaway: Filling a removable hole uses the limit value.

Authored solution to an added practice problem.

1. $\left|\right(3x-2)-(3a-2\left)\right|=3|x-a|$.

   Why: The vertical translation cancels.

2. If $0<|x-a|<\epsilon /3$, then the output error is less than ε.

   Why: The strict input inequality gives the strict output inequality.

Reviewed answer: $\delta =\epsilon /3$ works.

Takeaway: The input tolerance is the output tolerance divided by the absolute slope.

Authored solution to an added practice problem.

1. $|x^2-4|=|x-2||x+2|$.

   Why: Factor the output error.

2. Impose $|x-2|<1$, so $1<x<3$ and $|x+2|<5$.

   Why: A local bound turns the second factor into a fixed constant.

3. Also impose $|x-2|<\epsilon /5$ to get $|x^2-4|<5(\epsilon /5)=\epsilon$.

   Why: The minimum enforces both requirements at once.

Reviewed answer: $\delta =min(1,\epsilon /5)$ works.

Takeaway: For nonlinear errors, first bound the extra factor in a small neighborhood.

Authored solution to an added practice problem.

1. Rewrite $\sin \left(5x\right)/\left(2x\right)=(5/2)\sin \left(5x\right)/\left(5x\right)$.

   Why: Match the angle and denominator in the known sine limit.

2. As x→0, 5x→0 and the last ratio tends to 1.

   Why: The special limit uses radians.

Reviewed answer: $5/2$

Takeaway: Make the denominator match the sine's angle.

Authored solution to an added practice problem.

1. $1-\cos \left(3x\right)=2{\sin }^2(3x/2)$.

   Why: The half-angle identity exposes a square.

2. Divide by x² to get $(9/2)\left[\sin \right(3x/2)/(3x/2){]}^2$, which tends to 9/2.

   Why: Both sine ratios tend to 1.

Reviewed answer: $9/2$

Takeaway: A squared sine limit handles a cosine difference.

Authored solution to an added practice problem.

1. Since $-1\le \sin (1/x)\le 1$, multiplying by x² gives $-x^2\le x^2\sin (1/x)\le x^2$.

   Why: x² is nonnegative on both sides of zero.

2. Both bounds tend to 0, so the middle expression does too.

   Why: The squeeze theorem does not require the oscillating factor itself to have a limit.

Reviewed answer: $0$

Takeaway: A shrinking multiplier can control an oscillation.

Authored solution to an added practice problem.

1. Divide numerator and denominator by x² to obtain $(6-1/x)/(2/x^2-4)$.

   Why: The highest power removes the large common scale.

2. The reciprocal terms tend to 0, leaving 6/(−4)=−3/2.

   Why: The limiting denominator is nonzero.

Reviewed answer: $-3/2$

Takeaway: Keep the leading coefficient's sign.

Authored solution to an added practice problem.

1. Divide by x³: $(2/x^2+1/x^3)/(1-7/x^3)$.

   Why: Use the denominator's larger degree.

2. The numerator tends to 0 and denominator to 1.

   Why: The denominator grows two powers faster than the leading numerator term.

Reviewed answer: $0$

Takeaway: A lower-degree numerator is overwhelmed at infinity.

Authored solution to an added practice problem.

1. For x<2, x−2 is negative and tends to 0, so its reciprocal tends to −∞.

   Why: The denominator's sign controls the quotient.

2. For x>2, the reciprocal tends to +∞.

   Why: The opposite denominator sign gives the opposite one-sided behavior.

Reviewed answer: Left limit $-\infty$; right limit $+\infty$.

Takeaway: An infinite two-sided limit needs matching signs.

Authored solution to an added practice problem.

1. For x<0, $\left|x\right|=-x$, so $\left|x\right|/x=-1$.

   Why: The absolute-value formula depends on the side of zero.

2. For x>0, $\left|x\right|=x$, so the quotient is 1.

   Why: The values stay different however close x gets to zero.

Reviewed answer: Left limit −1; right limit 1; no two-sided limit.

Takeaway: A constant value on each side can still produce a jump.

Authored solution to an added practice problem.

1. The left entries 2.9 and 2.99 move toward 3; the right entries 3.01 and 3.1 are also compatible with approaching 3.

   Why: Use values nearest the target from both directions.

2. The table leaves all other nearby inputs unknown, so state a conjecture rather than a proof.

   Why: A limit describes every sufficiently close input, not just sampled ones.

Reviewed answer: The entries suggest a limit of 3; a finite table alone cannot prove it.

Takeaway: Tables give evidence for a limit, while a formula or bound can establish it.

Authored solution to an added practice problem.

1. The inner polynomial x²+1 tends to 5 as x→2.

   Why: Polynomials are continuous.

2. The positive square root is continuous at the positive input 5, so the composite tends to √5.

   Why: A continuous outer function carries the inner limit through the composition.

Reviewed answer: $\sqrt{5}$

Takeaway: Check that the outer function is continuous at the inner limit.

Authored solution to an added practice problem.

1. Expand $(3+h{)}^2-9=6h+h^2$.

   Why: The difference quotient compares nearby squared values.

2. For h≠0 divide by h to get $6+h$, whose limit is 6.

   Why: The derivative comes from a limit after cancellation.

Reviewed answer: $6$

Takeaway: An algebraic expansion can make a derivative limit straightforward.

Authored solution to an added practice problem.

1. Factor $x^2+x-2=(x-1)(x+2)$ and $x^2-1=(x-1)(x+1)$.

   Why: Both polynomials vanish at x=1.

2. The quotient is $(x+1)/(x+2)$ for x≠1, so the limit is 2/3.

   Why: Cancel only the common factor, leaving the distinct remaining factors.

Reviewed answer: $2/3$

Takeaway: Keep track of which factor remains above and below.

Authored solution to an added practice problem.

1. Both one-sided function limits of $\left|x\right|$ at zero are 0, equal to f(0).

   Why: That establishes continuity.

2. The quotient $\left[\right|h|-0]/h$ is −1 for negative h and 1 for positive h.

   Why: The derivative tests slopes, not just heights.

3. The slopes disagree, so the graph has a corner at zero.

   Why: Continuity alone does not guarantee a derivative.

Reviewed answer: Continuous at zero, but left slope −1 and right slope 1.

Takeaway: Meeting heights do not necessarily mean meeting slopes.

Authored solution to an added practice problem.

1. Match $(1+k/n{)}^n\to e^k$ with k=3.

   Why: This is the defining exponential limit from the source's special limits.

2. The limiting value is $e^3$.

   Why: The increment's coefficient becomes the exponential's exponent.

Reviewed answer: $e^3$

Takeaway: Identify the increment and the growing exponent together.