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Grade 12 Calculus · Lesson 1.390 min

Evaluating Limits Analytically

Evaluate limits exactly using substitution, limit laws, factoring, rationalizing, special trigonometric limits, and the Squeeze Theorem.

Choose a section to read or explore.

Lesson overview

In the previous lesson, you estimated limits from tables and graphs. Now we calculate them exactly. The central habit is simple: substitute first. If the result is a real number, stop. If the result is 0/00/0, that is a signal to rewrite the expression — never the final answer.

Explanation
Learning objectives

By the end of this lesson you should be able to:

  • Apply the limit laws and evaluate polynomial, rational, radical, composite, and trigonometric limits by direct substitution when valid.
  • Recognize that 0/00/0 is an indeterminate form, not an answer.
  • Rewrite an indeterminate quotient by factoring and cancelling a common factor.
  • Use the Factor Theorem, synthetic division, and polynomial long division when a polynomial factor is hidden.
  • Simplify complex rational expressions by building a common denominator.
  • Use a conjugate to rationalize a numerator or denominator containing radicals, and handle cube-root forms with the appropriate algebraic identity or substitution.
  • Compare left-hand and right-hand limits at piecewise, absolute-value, and greatest-integer breaking points.
  • Distinguish normal points from breaking points for greatest-integer functions.
  • Recognize endpoint/domain behavior and oscillating functions whose two-sided limit does not exist.
  • Apply the composite-function limit law when the outer function is continuous at the inner limit.
  • Determine a parameter by forcing the left-hand and right-hand limits to agree.
  • Evaluate trigonometric limits using lim⁡u→0sin⁡uu=1\lim_{u\to0}\frac{\sin u}{u}=1 and lim⁡u→01−cos⁡uu=0\lim_{u\to0}\frac{1-\cos u}{u}=0.
  • Use the Squeeze Theorem when direct algebra does not expose the limit.
Definition

Indeterminate form

The form 0/00/0 does not tell us the value of a limit. It tells us that the numerator and denominator share behavior that may be hidden. Rewrite the function for nearby inputs, then evaluate the equivalent expression.

Mr. Azzam’s limit decision routine

Use this routine on every analytical limit. It keeps the work short and prevents you from choosing a complicated method too early.

Step 01

Substitute

Insert the target value before doing any algebra.

Step 02

Read the result

A finite number is the answer. A nonzero denominator is safe.

Step 03Do something

If 0/00/0: do something

Factor and cancel, rationalize, or use a trigonometric identity.

Step 04

Substitute again

Evaluate the simpler nearby expression and check the result.

Method chooser

Polynomial factors?
Factor and cancel.
Radicals?
Multiply by the conjugate.
Sine or cosine near zero?
Build a special limit.

Your turn

Direct substitution in a rational expression gives 0/00/0. What does this tell you?

Direct substitution and limit laws

Polynomials are continuous everywhere. Rational functions are continuous wherever their denominators are nonzero. Radicals and composite functions may also be evaluated by substitution when the substituted value lies in the domain.

lim⁡(f(x)+g(x))\lim(f(x)+g(x))becomesL+ML+M
lim⁡(f(x)−g(x))\lim(f(x)-g(x))becomesL−ML-M
lim⁡(cf(x))\lim(c f(x))becomescLcL
lim⁡(f(x)g(x))\lim(f(x)g(x))becomesLMLM
lim⁡f(x)g(x)\lim\frac{f(x)}{g(x)}becomesLM\frac{L}{M}, provided M≠0M\ne0
lim⁡[f(x)]n\lim[f(x)]^nbecomesLnL^n
Example 1 · Substitute and stop
lim⁡x→3(4x2−5x+2)\displaystyle \lim_{x\to3}(4x^2-5x+2)

Try each checkpoint, or reveal all worked steps whenever you need them.

Your turn · Step 1 · Decide

What should you do first?

Example 2 · Check the denominator
lim⁡x→2x2+3x+1\displaystyle \lim_{x\to2}\frac{x^2+3}{x+1}

Try each checkpoint, or reveal all worked steps whenever you need them.

Your turn · Step 1 · Check the denominator

What does the denominator approach?

Common mistake. Do not factor automatically. If direct substitution gives a real number, extra algebra wastes time and creates more chances for an error.

Your turn

Evaluate lim⁡x→−2(3x2−4x+1)\displaystyle \lim_{x\to -2}(3x^2-4x+1).

Dividing out: factor and cancel

When substitution gives 0/00/0 in a rational expression, look for a common factor. Cancelling is valid for nearby values where x≠cx\ne c. A limit describes those nearby values, not the missing point itself.

a2−b2=(a−b)(a+b)a3−b3=(a−b)(a2+ab+b2)a^2-b^2=(a-b)(a+b)\qquad a^3-b^3=(a-b)(a^2+ab+b^2)
Example 3 · Difference of squares
lim⁡x→3x2−9x−3\displaystyle \lim_{x\to3}\frac{x^2-9}{x-3}

Try each checkpoint, or reveal all worked steps whenever you need them.

Your turn · Step 1 · Substitute first

Direct substitution gives:

Example 4 · Factor both numerator and denominator
lim⁡x→1x2+x−2x2−1\displaystyle \lim_{x\to1}\frac{x^2+x-2}{x^2-1}

Try each checkpoint, or reveal all worked steps whenever you need them.

Your turn · Step 1 · Diagnose

What does direct substitution produce?

Common mistake. Cancel factors, not terms. In (x2−9)/(x−3)(x^2-9)/(x-3), you must factor the numerator before cancelling.

Your turn

What is the first useful rewriting step for lim⁡x→3x2−9x−3\displaystyle \lim_{x\to3}\frac{x^2-9}{x-3}?

Factor Theorem, synthetic division, and long division

When substitution gives 0/00/0 but a polynomial factor is hidden, use the Factor Theorem. If P(c)=0P(c)=0, then x−cx-c is a factor. Synthetic division is efficient for a divisor of the form x−cx-c; polynomial long division handles a general polynomial divisor.

P(x)=D(x)Q(x)+R(x)P(x)D(x)=Q(x)+R(x)D(x)P(x)=D(x)Q(x)+R(x)\qquad\frac{P(x)}{D(x)}=Q(x)+\frac{R(x)}{D(x)}
Interactive example · Synthetic division
lim⁡x→3x3−7x−6x−3\displaystyle \lim_{x\to3}\frac{x^3-7x-6}{x-3}

Try each checkpoint, or reveal all worked steps whenever you need them.

Your turn · Step 1 · Substitute first

What does direct substitution give?

Interactive example · Polynomial long division
lim⁡x→2x3+x2−4x−4x2−4\displaystyle \lim_{x\to2}\frac{x^3+x^2-4x-4}{x^2-4}

Try each checkpoint, or reveal all worked steps whenever you need them.

Your turn · Step 1 · Substitute first

What does direct substitution give?

Common mistake. Synthetic division applies directly to a linear divisor of the form x−cx-c. Do not force it onto a higher-degree divisor.

Your turn

For lim⁡x→3x3−7x−6x−3\displaystyle\lim_{x\to3}\frac{x^3-7x-6}{x-3}, what quotient remains after exact polynomial division?

Complex rational expressions

When smaller fractions create the indeterminate form, first build a common denominator. This exposes the factor that can be cancelled.

Interactive example · Build a common denominator
lim⁡x→21x−12x−2\displaystyle \lim_{x\to2}\frac{\frac1x-\frac12}{x-2}

Try each checkpoint, or reveal all worked steps whenever you need them.

Your turn · Step 1 · Substitute first

What does direct substitution give?

Your turn

For lim⁡x→21x−12x−2\displaystyle\lim_{x\to2}\frac{\frac1x-\frac12}{x-2}, which simplified expression agrees for nearby x≠2x\ne2?

Rationalizing technique

If 0/00/0 comes from a square root, multiply by the conjugate. The product(a−b)(a+b)=a2−b2(a-b)(a+b)=a^2-b^2 removes the radical difference and usually reveals a factor that cancels.

Conjugate of u−a is u+a.\text{Conjugate of }\sqrt{u}-a\text{ is }\sqrt{u}+a.
Example 5 · Rationalize the numerator
lim⁡x→4x+5−3x−4\displaystyle \lim_{x\to4}\frac{\sqrt{x+5}-3}{x-4}

Try each checkpoint, or reveal all worked steps whenever you need them.

Your turn · Step 1 · Diagnose

Which technique exposes a cancellable factor?

Common mistake. When multiplying by the conjugate, multiply both the numerator and denominator. You are multiplying by 11, so the value of the expression does not change.

Your turn

For lim⁡x→4x+5−3x−4\displaystyle \lim_{x\to4}\frac{\sqrt{x+5}-3}{x-4}, which conjugate should be used?

Radical extensions: cube roots and substitution

The class notes extend the same routine beyond square-root conjugates. For cube-root differences, use the difference-of-cubes identity. When several fractional powers appear, introduce a substitution that converts them to ordinary polynomial powers.

a3−b3=(a−b)(a2+ab+b2)a^3-b^3=(a-b)(a^2+ab+b^2)
Interactive example · Cube-root difference
lim⁡x→8x3−2x−8\displaystyle \lim_{x\to8}\frac{\sqrt[3]{x}-2}{x-8}

Try each checkpoint, or reveal all worked steps whenever you need them.

Your turn · Step 1 · Substitute first

What does direct substitution give?

Interactive example · Fractional-power substitution
lim⁡x→1x2/3−1x1/2−1\displaystyle \lim_{x\to1}\frac{x^{2/3}-1}{x^{1/2}-1}

Try each checkpoint, or reveal all worked steps whenever you need them.

Your turn · Step 1 · Substitute first

What does direct substitution give?

Your turn

Question 1 of 2
For lim⁡x→1x2/3−1x1/2−1\displaystyle\lim_{x\to1}\frac{x^{2/3}-1}{x^{1/2}-1}, which substitution converts both powers to integer powers?

One-sided, piecewise, absolute value, and greatest-integer limits

At a breaking point, evaluate the left-hand and right-hand limits separately. A two-sided limit exists only when they agree.

lim⁡x→cf(x)=L⟺lim⁡x→c−f(x)=lim⁡x→c+f(x)=L\lim_{x\to c}f(x)=L\quad\Longleftrightarrow\quad\lim_{x\to c^-}f(x)=\lim_{x\to c^+}f(x)=L
Interactive example · Test a breaking point
f(x)={x+1,x<25−x,x≥2Find lim⁡x→2f(x).f(x)=\begin{cases}x+1,&x<2\\5-x,&x\ge2\end{cases}\qquad \text{Find }\lim_{x\to2}f(x).

Try each checkpoint, or reveal all worked steps whenever you need them.

Your turn · Step 1 · Left-hand limit

What is the left-hand limit?

For absolute value, rewrite around the sign-change point. For the greatest-integer function, identify the integer breaking points and use the correct subinterval on each side.

Your turn

For f(x)={x+1,x<25−x,x≥2f(x)=\begin{cases}x+1,&x<2\\5-x,&x\ge2\end{cases}, which pair establishes the limit at 22?

Absolute value and greatest-integer limits

For absolute value, rewrite the expression on each side of its sign-change point. For a greatest-integer function, substitute into the inside first: an integer result is a breaking point; otherwise it is a normal point.

∣u∣={u,u≥0−u,u<0|u|=\begin{cases}u,&u\ge0\\-u,&u<0\end{cases}
Class-note example · Absolute value
lim⁡x→0∣x∣x\displaystyle \lim_{x\to0}\frac{|x|}{x}

Try each checkpoint, or reveal all worked steps whenever you need them.

Your turn · Step 1 · Left side

Work this step on paper first. The solution stays hidden until you confirm you are ready.

Class-note routine · Greatest integer
lim⁡x→−2⌊x+3⌋\displaystyle \lim_{x\to-2}\lfloor x+3\rfloor

Try each checkpoint, or reveal all worked steps whenever you need them.

Your turn · Step 1 · Test the inside

Work this step on paper first. The solution stays hidden until you confirm you are ready.

Common mistake. Do not assume every target is a breaking point. Substitute into the inside expression first.

Your turn

For lim⁡x→−2⌊x+3⌋\displaystyle\lim_{x\to-2}\lfloor x+3\rfloor, what happens on the two sides?

Endpoint, domain, and oscillating limits

Check whether the real domain allows an approach from both sides. A function can also fail to have a limit because it keeps oscillating without settling at one value.

Endpoint example
lim⁡x→0x\displaystyle \lim_{x\to0}\sqrt{x}

Try each checkpoint, or reveal all worked steps whenever you need them.

Your turn · Step 1 · Domain

Work this step on paper first. The solution stays hidden until you confirm you are ready.

Oscillation example
lim⁡x→0sin⁡(1x)\displaystyle \lim_{x\to0}\sin\left(\frac1x\right)

Try each checkpoint, or reveal all worked steps whenever you need them.

Your turn · Step 1 · Inspect the input

Work this step on paper first. The solution stays hidden until you confirm you are ready.

Interactive lab

Oscillation near zero

Zoom toward zero, then compare two sequences of positive inputs approaching zero. Do their outputs settle to the same value?
f(x)=sin⁡(1/x),x≠0f(x)=\sin(1/x),\quad x\ne0
-101−0.3500.35xy

Independent axes. Shaded |x| < 0.00393 is unresolved, not blank function values: infinitely many oscillations continue there. x = 0 is undefined.

Optional Desmos exploration

The native graph and checks work without this external service.

xₙ = 0.070736 → output exactly 1

zₙ = 0.057875 → output exactly −1

Increase n: both inputs approach 0 from the right.

What can you conclude about the limit as x → 0?

Your turn

Why does lim⁡x→0sin⁡(1/x)\displaystyle\lim_{x\to0}\sin(1/x) fail to exist?

Limit of a composite function

lim⁡x→ag(x)=L,f continuous at L⟹lim⁡x→af(g(x))=f(L)\lim_{x\to a}g(x)=L,\quad f\text{ continuous at }L\quad\Longrightarrow\quad\lim_{x\to a}f(g(x))=f(L)
Composite-function example
lim⁡x→2x2+5\displaystyle \lim_{x\to2}\sqrt{x^2+5}

Try each checkpoint, or reveal all worked steps whenever you need them.

Your turn · Step 1 · Inner limit

Work this step on paper first. The solution stays hidden until you confirm you are ready.

Your turn

If g(x)→9g(x)\to9 as x→2x\to2, which fact justifies g(x)→3\sqrt{g(x)}\to3?

Special trigonometric limits

These limits are valid when angles are measured in radians. Your goal is to reshape the expression until the angle in the sine matches its denominator.

lim⁡u→0sin⁡uu=1andlim⁡u→01−cos⁡uu=0\lim_{u\to0}\frac{\sin u}{u}=1\qquad\text{and}\qquad\lim_{u\to0}\frac{1-\cos u}{u}=0
Example 6 · Build the standard form
lim⁡x→0sin⁡(5x)x\displaystyle \lim_{x\to0}\frac{\sin(5x)}{x}

Try each checkpoint, or reveal all worked steps whenever you need them.

Your turn · Step 1 · Match angle and denominator

Which factor must be introduced?

Example 7 · Two special-limit factors
lim⁡x→0sin⁡(3x)sin⁡(2x)\displaystyle \lim_{x\to0}\frac{\sin(3x)}{\sin(2x)}

Try each checkpoint, or reveal all worked steps whenever you need them.

Your turn · Step 1 · Build two known forms

Which constant remains after matching both angles?

Common mistake. The angle must approach 00, and the formula uses radians. Do not applysin⁡u/u→1\sin u/u\to1 when the denominator does not match the angle until you insert the required constant factor.

Your turn

Evaluate lim⁡x→0sin⁡(5x)x\displaystyle \lim_{x\to0}\frac{\sin(5x)}{x} in radians.

Trigonometric identities and extended special limits

The notes build harder trigonometric limits from identities and the basic special limits. The method is to rewrite first, then create sine-over-angle or tangent-over-angle factors.

lim⁡x→0sin⁡(ax)bx=ablim⁡x→0tan⁡(ax)bx=ab\lim_{x\to0}\frac{\sin(ax)}{bx}=\frac ab\qquad\lim_{x\to0}\frac{\tan(ax)}{bx}=\frac ab
Interactive example · Mixed trigonometric ratio
lim⁡x→0sin⁡(4x)tan⁡(3x)\displaystyle \lim_{x\to0}\frac{\sin(4x)}{\tan(3x)}

Try each checkpoint, or reveal all worked steps whenever you need them.

Your turn · Step 1 · Build known forms

Which constant factor remains after matching the angles?

Interactive example · Identity before special limit
lim⁡x→01−cos⁡(2x)x2\displaystyle \lim_{x\to0}\frac{1-\cos(2x)}{x^2}

Try each checkpoint, or reveal all worked steps whenever you need them.

Your turn · Step 1 · Use an identity

Which identity is useful first?

Your turn

Evaluate lim⁡x→01−cos⁡(2x)x2\displaystyle\lim_{x\to0}\frac{1-\cos(2x)}{x^2} in radians.

Parameter and difference-quotient limits

Later examples in the notes reverse the process: determine a parameter so the required limit exists, or simplify a difference-quotient form before taking the limit.

Interactive example · Difference quotient structure
lim⁡h→0(x+h)2−x2h\displaystyle \lim_{h\to0}\frac{(x+h)^2-x^2}{h}

Try each checkpoint, or reveal all worked steps whenever you need them.

Your turn · Step 1 · Substitute first

What does direct substitution produce?

Common mistake. A parameter question is not solved by guessing the parameter. Use the condition required for the relevant left/right expressions or algebraic cancellation, then solve that condition.

Your turn

For fixed xx, simplify (x+h)2−x2h\displaystyle\frac{(x+h)^2-x^2}{h} when h≠0h\ne0.

The Squeeze Theorem

If a difficult function stays between two easier functions and both outside functions approach the same number, the middle function must approach that number too.

g(x)≤f(x)≤h(x),lim⁡x→cg(x)=lim⁡x→ch(x)=L⟹lim⁡x→cf(x)=Lg(x)\le f(x)\le h(x),\quad \lim_{x\to c}g(x)=\lim_{x\to c}h(x)=L\quad\Longrightarrow\quad\lim_{x\to c}f(x)=L
Example 8 · Squeeze an oscillating factor
lim⁡x→0x2cos⁡(1x)\displaystyle \lim_{x\to0}x^2\cos\left(\frac1x\right)

Try each checkpoint, or reveal all worked steps whenever you need them.

Your turn · Step 1 · Find reliable bounds

Which bounds both trap the function and approach the same limit at zero?

Check 1

Find reliable bounds.

Check 2

Evaluate both outside limits.

Check 3

If both equal LL, conclude the middle limit is LL.

Your turn

If −x2≤f(x)≤x2-x^2\le f(x)\le x^2 near x=0x=0, what is lim⁡x→0f(x)\displaystyle \lim_{x\to0}f(x)?

Study resources

Use the class notes and textbook as your main sequence. The external resources provide another explanation, additional worked examples, video practice, and visual checking.

Primary class source

Mr. Ahmed Azzam's Lesson 1.3 notes

The lesson sequence, colour-coded method choice, and worked-example style used on this page.

Primary textbook source

Evaluating Limits Analytically · Textbook 1.3

Limit laws, direct substitution, dividing out, rationalizing, the Squeeze Theorem, and trigonometric limits.

Read and practise

OpenStax Calculus · The Limit Laws

A second explanation with examples on factoring, conjugates, and the Squeeze Theorem.

More worked examples

Paul's Online Notes · Computing Limits

Extra examples for choosing and applying an algebraic limit technique.

Video and practice

Khan Academy · Limits and continuity

Short lessons and practice for reviewing limit ideas at your own pace.

Visual verification

Desmos Graphing Calculator

Graph the original and simplified expressions to see their nearby behaviour and any removable hole.

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