Modules › Limits
Limits
The limit idea, ε–δ definition, limit laws, one-sided limits, squeeze theorem, infinite limits and limits at infinity.
Lectures 1–2 · Tutorial 2 · Functions, limits and continuity
Textbook sections: Thomas ch. 1 Functions (1.1–1.3) and ch. 2 Limits and Continuity (2.1–2.6) · Stewart ch. 1 (1.1–1.3) and ch. 2 (2.1–2.6)
Textbook sections: Thomas ch. 1 Functions (1.1–1.3) and ch. 2 Limits and Continuity (2.1–2.6) · Stewart ch. 1 (1.1–1.3) and ch. 2 (2.1–2.6)
Limits
Calculus I · Module 2
After this module you can
- State the ε–δ definition and choose δ for a linear function.
- Evaluate limits by factoring, conjugates, dividing by the highest power, the standard limit $\sin u/u\to1$ and the squeeze theorem.
- Compute one-sided limits and decide when a limit does not exist.
- Evaluate infinite limits and limits at infinity (watch the sign for $x\to-\infty$).
What a limit says
- $\lim_{x\to a}f(x)=L$: $f(x)$ gets as close to $L$ as we like when $x$ is close to $a$, $x\ne a$. The value $f(a)$ is irrelevant.
- ε–δ: for every $\varepsilon>0$ there is $\delta>0$ with $0<|x-a|<\delta\Rightarrow|f(x)-L|<\varepsilon$.
- For $f(x)=mx+c$: $|f(x)-L|=|m||x-a|$, so $\delta=\varepsilon/|m|$ works.
- Limit exists iff both one-sided limits exist and are equal.
Limit laws and the standard toolbox
| Situation | Technique |
|---|---|
| $\frac00$ with polynomials | factor and cancel |
| $\frac00$ with roots | multiply by the conjugate |
| $x\to\pm\infty$, rational | divide by the highest power of $x$ |
| trigonometric | $\frac{\sin u}{u}\to1$, $\frac{1-\cos u}{u^2}\to\frac12$ |
| oscillating × small | squeeze: $-|g|\le f\le|g|$ |
Squeeze theorem and one-sided limits
- If $g\le f\le h$ near $a$ and $g,h\to L$, then $f\to L$.
- Example: $-x^2\le x^2\sin\frac1x\le x^2$, so $\lim_{x\to0}x^2\sin\frac1x=0$.
- $\lim_{x\to0}\frac{|x|}{x}$ does not exist: left limit $-1$, right limit $+1$.
- Piecewise functions: compute each one-sided limit with the matching formula.
Infinite limits and limits at infinity
- $\lim_{x\to1^+}\frac{1}{x-1}=+\infty$ and $\lim_{x\to1^-}\frac1{x-1}=-\infty$: a vertical asymptote $x=1$.
- $\lim_{x\to\infty}\frac{3x^2-x}{2x^2+5}=\frac32$: horizontal asymptote $y=\frac32$.
- $\lim_{x\to-\infty}\frac{x}{\sqrt{x^2+1}}=-1$ because $\sqrt{x^2}=|x|=-x$ for $x<0$.
- $\lim_{x\to\infty}(\sqrt{x^2+x}-x)=\frac12$: multiply by the conjugate.
A model ε–δ proof
- Claim: $\lim_{x\to4}(9-x)=5$. Given $\varepsilon>0$ choose $\delta=\varepsilon$.
- If $0<|x-4|<\delta$ then $|(9-x)-5|=|4-x|<\varepsilon$. ∎
- For $\frac{x^2-9}{x+3}\to-6$ at $x=-3$: for $x\ne-3$ the quotient equals $x-3$, so again $\delta=\varepsilon$.
- For non-linear $f$ restrict $\delta\le1$ first to bound the extra factor (e.g. $|x+2|\le5$ near $x=2$).
Typical mistakes
Substituting too early$\frac00$ is not $0$ or $1$: factor first.
Sign at $-\infty$$\sqrt{x^2}=|x|$, not $x$.
'It does not exist' too fast$\lim x^2\sin\frac1x=0$ even though $\sin\frac1x$ oscillates.
Practice next
- Chapter quizzes Ch. 1 (3 sets)
- Practice bank, filtered to this chapter, then written mode
- Lab: Limit explorer
- Then the mock exams (Exams page)
Worked examples from the course problems
Worked example 1 · Epsilon–delta (Tutorial 2, Q6(a))
The exit rate is $y=\sqrt{x}/2$ ft³/min for water depth $x$ ft. To keep $y$ within $0.2$ of $y_0=1$, the depth must stay within … ft of $x_0=4$
Show the answer and the solution
Answer: $1.44$
$0.8<\sqrt x/2<1.2\iff2.56<x<5.76$. The nearer endpoint to $4$ is $2.56$, at distance $1.44$ (the other is $1.76$ away), so $\delta=1.44$.
Worked example 2 · Limits (Mock Test 1, Q1(b)(i))
Evaluate $\displaystyle\lim_{x\to\infty}\dfrac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{\sqrt{x}+1}$.
Show the answer and the solution
Answer: $1$
Divide by $\sqrt x$: numerator $\sqrt{1+\sqrt{1/x+1/x^{3/2}}}\to1$, denominator $1+1/\sqrt x\to1$.
Worked example 3 · Epsilon–delta (Tutorial 2, Q6(b))
With the same exit rate $y=\sqrt x/2$, to keep $y$ within $0.1$ of $1$ the depth must stay within … ft of $4$
Show the answer and the solution
Answer: $0.76$
$0.9<\sqrt x/2<1.1\iff3.24<x<4.84$. Distances from $4$: $0.76$ and $0.84$; take the smaller, $\delta=0.76$.
Worked example 4 · Limits (Mock Test 1, Q1(b)(ii))
Evaluate $\displaystyle\lim_{x\to\infty}\dfrac{e^{\sin x}}{x+1}$.
Show the answer and the solution
Answer: $0$
Squeeze: $\frac{1}{e}\le e^{\sin x}\le e$, so $0<\frac{e^{\sin x}}{x+1}\le\frac{e}{x+1}\to0$.
Practice
Quiz set A Quiz set B Quiz set C Practice bank (this chapter) Written mode Lab: Limit explorer