Computing Limits with Reasons

Substitution is often the last line of a limit calculation, but it is not the definition of a limit. It works for polynomials because their limits obey algebraic rules; it works for a rational expression at points where the denominator has a nonzero limit. Recognizing those conditions lets us distinguish an ordinary calculation from one that needs more work.

The algebra of finite limits

Suppose $\lim_{x\to a}f(x)=L$ and $\lim_{x\to a}g(x)=M$, with $L,M$ finite. Then limits commute with sums, differences, constant multiples, and products. Thus $\lim(f+g)=L+M$ and $\lim(fg)=LM$. For quotients we additionally require $M\ne0$, giving $\lim(f/g)=L/M$. Positive integer powers follow by repeated products.

For composition, if $f(x)\to L$ and $F$ is continuous at $L$, then $F(f(x))\to F(L)$, provided the composition is defined nearby. Continuity will receive a careful definition shortly. Root functions are continuous on their real domains, with one-sided care at endpoints. Hence

$$\lim_{x\to2}\frac{3x^2-x+1}{x+4}=\frac{11}{6}.$$

The denominator approaches $6$, not zero. That single check justifies the final division.

Zero over zero is a prompt to investigate

For $(x^2-9)/(x-3)$ at $3$, substitution gives the meaningless quotient $0/0$. It does not give an answer of zero, infinity, or “undefined limit.” For $x\ne3$,

$$\frac{x^2-9}{x-3}=x+3,$$

so the limit is $6$. The two expressions agree throughout a punctured neighborhood. Limits depend on exactly that nearby behavior.

Compare $x/x$, $x^2/x$, and $x/x^2$ at zero. All display $0/0$ on substitution, yet their behaviors are respectively a constant $1$, a limit $0$, and incompatible unbounded one-sided limits. The notation indeterminate form records insufficient information, not a result.

Conjugates handle some square-root cases. For $x\ne4$ near $4$,

$$\frac{\sqrt{x}-2}{x-4} =\frac1{\sqrt{x}+2},$$

because multiplying numerator and denominator by $\sqrt{x}+2$ exposes $x-4$. The limit is $1/4$. Multiplying by the conjugate is legal nearby because $\sqrt{x}+2$ is nonzero.

Squeezing every nearby value

The squeeze theorem says that if $u(x)\leq f(x)\leq v(x)$ for all sufficiently near allowed inputs, and both bounding functions approach $L$, then $f(x)$ also approaches $L$. The inequalities must hold throughout the relevant neighborhood, not only at selected table entries.

Since $|\sin(1/x)|\leq1$, we have

$$-|x|\leq x\sin(1/x)\leq |x|.$$

Both outer expressions tend to zero. This supplies a conclusion even though the middle function oscillates infinitely often near zero.

An important trigonometric limit is $\lim_{x\to0}\sin x/x=1$, with angles measured in radians. On $0<x<\pi/2$, the unit-circle area comparison gives $\sin x<x<\tan x$. Dividing yields $\cos x<\sin x/x<1$. As $x\to0^+$, both bounds tend to $1$; symmetry gives the left side. This argument uses elementary geometric continuity of cosine, not a derivative formula for sine. Also,

$$\frac{1-\cos x}{x}=\frac{\sin x}{x}\frac{\sin x}{1+\cos x}\longrightarrow0.$$

These results will justify trigonometric derivatives. Degrees would introduce a conversion factor; the radian convention is mathematical infrastructure, not a cosmetic preference.

Unbounded inputs and dominant terms

To study a rational expression as $x\to\infty$, divide numerator and denominator by a suitable highest power. For example,

$$\frac{5x^2+2x-1}{2x^2+7} =\frac{5+2/x-1/x^2}{2+7/x^2}\longrightarrow\frac52.$$

If the denominator degree is larger, the ratio tends to zero. If the numerator degree is larger, polynomial growth remains; its sign and direction must still be checked. A memorized degree rule should summarize this calculation rather than replace it.

For radicals, $\sqrt{x^2}=|x|$, not always $x$. Therefore $\sqrt{x^2+1}/x$ tends to $1$ as $x\to+\infty$ and $-1$ as $x\to-\infty$. Forgetting the absolute value changes the second answer.

Exercises

1. Evaluate $\lim_{x\to2}(x^2+x-6)/(x-2)$.
2. Evaluate $\lim_{x\to0}(\sqrt{1+x}-1)/x$.
3. Find $\lim_{x\to0}\sin(5x)/x$ and justify the factor $5$.
4. Use squeezing to evaluate $\lim_{x\to0}x^2\cos(1/x)$.
5. Find the limits of $(3x^3-x)/(x^3+2)$ at positive and negative infinity.
6. Explain why the quotient law cannot establish $\lim_{x\to0}\sin x/x$ directly.
7. Find $\lim_{x\to-\infty}\sqrt{4x^2+1}/x$.
8. Give an example showing that $\lim(fg)$ can exist even when neither factor has a finite limit.

Worked solutions

1. Factor the numerator as $(x+3)(x-2)$; the limit is $5$.
2. The conjugate gives $1/(\sqrt{1+x}+1)$ near zero, with limit $1/2$.
3. Write $5[\sin(5x)/(5x)]$. The bracket tends to $1$ in radians, giving $5$.
4. The absolute value is bounded by $x^2$, so the limit is zero.
5. Dividing by $x^3$ gives $3$ in either direction. The nonzero denominator limit justifies the quotient step.
6. The denominator limit is zero, violating that law’s nonzero-denominator hypothesis.
7. Write $|x|\sqrt{4+1/x^2}/x$. For negative $x$, $|x|/x=-1$, giving $-2$.
8. Let $f(x)=1/x$ and $g(x)=x$ for $x\ne0$. Their product has limit $1$ at zero, although the first factor has no finite limit there. To make neither factor converge, take $f(x)=2+\sin(1/x)$ and $g(x)=1/f(x)$; their product is identically $1$ while both factors oscillate.