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๐ŸŽฏโญ INTERACTIVE LESSON

Direct Substitution Method

Learn step-by-step with interactive practice!

Direct Substitution Method - Complete Interactive Lesson

Part 1: The Idea of Direct Substitution

๐ŸŽฏ Direct Substitution โ€” The Idea

Part 1 of 4 โ€” When you can "just plug in"


Topics in This Part

Section
The Big Idea
๐Ÿ”‘ The Continuity Connection
Easy Examples

๐Ÿ”‘ Why this matters: Direct substitution is the first thing you try on every limit problem. Many AP limits are solved in one line.

๐Ÿ’ก The Big Idea

Direct substitution: to find limโกxโ†’af(x)\lim_{x \to a} f(x), simply compute f(a)f(a).

If ff is continuous at aa, the answer is just f(a)f(a) โ€” no fancy work needed.

limโกxโ†’af(x)=f(a)wheneverย fย isย continuousย atย a.\lim_{x \to a} f(x) = f(a) \quad \text{whenever $f$ is continuous at $a$.}

๐Ÿ’ก This is the definition of continuity: limit equals value.

๐Ÿ”‘ The Continuity Connection

Direct substitution works because of a chain of facts:

  1. Polynomials are continuous everywhere.
  2. Rational functions are continuous everywhere their denominator is nonzero.
  3. sinโกx\sin x, cosโกx\cos x, exe^x are continuous everywhere.
  4. lnโกx\ln x is continuous on x>0x > 0.
  5. x\sqrt{x} is continuous on xโ‰ฅ0x \ge 0.
  6. Sums, products, quotients (denominator โ‰ 0\ne 0), and compositions of continuous functions are continuous.

๐Ÿ’ก Strategy: before trying anything fancy, try plugging in. If it gives a clean number, you're done.

๐Ÿ“ Easy Examples

LimitSubstitutionAnswer
limโกxโ†’4(3x2โˆ’2x+1)\lim_{x \to 4} (3x^2 - 2x + 1)3(16)โˆ’2(4)+1=413(16) - 2(4) + 1 = 414141
limโกxโ†’0cosโกx\lim_{x \to 0} \cos xcosโก0=1\cos 0 = 111
limโกxโ†’1x2+5x+2\lim_{x \to 1} \dfrac{x^2 + 5}{x + 2}1+51+2=2\dfrac{1 + 5}{1 + 2} = 222
limโกxโ†’elnโกx\lim_{x \to e} \ln xlnโกe=1\ln e = 111
limโกxโ†’2ex\lim_{x \to 2} e^xe2โ‰ˆ7.389e^2 \approx 7.389e2e^2

Concept Check ๐ŸŽฏ

Substitute ๐Ÿงฎ

Compute each limit by direct substitution:

1) limโกxโ†’2(x3โˆ’x)=?\lim_{x \to 2} (x^3 - x) = ?

2) limโกxโ†’0(sinโกx+5)=?\lim_{x \to 0} (\sin x + 5) = ?

3) limโกxโ†’4x+1xโˆ’1=?\lim_{x \to 4} \dfrac{x + 1}{x - 1} = ? (decimal or fraction)

Part 2: When Direct Substitution Works

โœ… When Direct Substitution Works

Part 2 of 4 โ€” A continuity catalog


Topics in This Part

Section
Polynomials
Rational Functions
Trig, Exp, Log, Roots

๐Ÿ”‘ Why this matters: Knowing the "continuity catalog" lets you spot a one-line problem at a glance.

๐Ÿ“ Polynomials

A polynomial p(x)=anxn+โ‹ฏ+a0p(x) = a_n x^n + \cdots + a_0 is continuous everywhere.

So: limโกxโ†’ap(x)=p(a)forย anyย realย a.\lim_{x \to a} p(x) = p(a) \quad \text{for any real } a.

๐Ÿ’ก Always try substitution first for polynomials. It will work.

รท Rational Functions

A rational function r(x)=p(x)q(x)r(x) = \dfrac{p(x)}{q(x)} is continuous wherever q(x)โ‰ 0q(x) \ne 0.

SituationResult
q(a)โ‰ 0q(a) \ne 0Direct substitution gives the limit immediately
q(a)=0q(a) = 0 but p(a)โ‰ 0p(a) \ne 0Limit DNE (vertical asymptote) โ€” see Infinite Limits
q(a)=0q(a) = 0 AND p(a)=0p(a) = 0Indeterminate 0/00/0 โ€” try factoring/rationalizing/L'Hรดpital

Worked example: limโกxโ†’5x2โˆ’1xโˆ’2=243=8\lim_{x \to 5} \dfrac{x^2 - 1}{x - 2} = \dfrac{24}{3} = 8. โœ…

๐ŸŒ€ Trig, Exp, Log, Roots

FunctionContinuous on
sinโกx\sin x, cosโกx\cos xAll of R\mathbb{R}
exe^xAll of R\mathbb{R}
lnโกx\ln xx>0x > 0
x\sqrt{x}xโ‰ฅ0x \ge 0
tanโกx\tan xAll xx except odd multiples of ฯ€/2\pi/2
secโกx\sec x, cscโกx\csc x, cotโกx\cot xWherever denominators don't vanish

โš ๏ธ For tanโกx\tan x at ฯ€/2\pi/2, x\sqrt{x} at negative numbers, and lnโกx\ln x at 0, direct substitution fails because the function isn't continuous (or even defined) there.

Catalog Check ๐ŸŽฏ

Substitute or Stop? ๐Ÿ”ฝ

Part 3: Limit Laws

๐Ÿ“œ Limit Laws โ€” The Algebra of Substitution

Part 3 of 4 โ€” Why "just plug in" works


Topics in This Part

Section
๐Ÿ”‘ The Five Core Limit Laws
Composition of Continuous Functions
Examples

๐Ÿ”‘ Why this matters: Limit laws are what justify direct substitution and let you break complex limits into pieces.

๐Ÿ”‘ The Five Core Limit Laws

If limโกxโ†’af(x)=L\lim_{x \to a} f(x) = L and limโกxโ†’ag(x)=M\lim_{x \to a} g(x) = M (both exist), and cc is a constant:

LawStatement
Sumlimโก(f+g)=L+M\lim (f + g) = L + M
Constant Multiplelimโก(cโ‹…f)=cL\lim (c \cdot f) = cL
Productlimโก(fg)=LM\lim (fg) = LM
Quotientlimโกfg=LM\lim \dfrac{f}{g} = \dfrac{L}{M} if Mโ‰ 0M \ne 0
Powerlimโกfn=Ln\lim f^n = L^n for positive integers nn

๐Ÿ’ก These laws let you take a complicated expression apart, evaluate each piece, and combine โ€” exactly what direct substitution does in shorthand.

๐Ÿ” Composition of Continuous Functions

If gg is continuous at aa and ff is continuous at g(a)g(a), then limโกxโ†’af(g(x))=f(g(a)).\lim_{x \to a} f(g(x)) = f(g(a)).

In words: "you can move the limit inside a continuous function."

Worked example: limโกxโ†’3x2+7\lim_{x \to 3} \sqrt{x^2 + 7}.

  • Inside: limโกxโ†’3(x2+7)=16\lim_{x \to 3} (x^2 + 7) = 16.
  • Outside: โ€‰โ‹…โ€‰\sqrt{\,\cdot\,} is continuous at 16.
  • Conclusion: 16=4\sqrt{16} = 4.

๐Ÿ“ Examples Combining the Laws

Ex 1. limโกxโ†’2[(x3โˆ’1)(x+5)]=(8โˆ’1)(7)=49\lim_{x \to 2} \bigl[(x^3 - 1)(x + 5)\bigr] = (8 - 1)(7) = 49. (Product law.)

Ex 2. limโกxโ†’0ex+4cosโกx+2=1+41+2=53\lim_{x \to 0} \dfrac{e^x + 4}{\cos x + 2} = \dfrac{1 + 4}{1 + 2} = \dfrac{5}{3}. (Quotient law.)

Ex 3. limโกxโ†’ฯ€/2sinโก(cosโกx)=sinโก(cosโก(ฯ€/2))=sinโก0=0\lim_{x \to \pi/2} \sin(\cos x) = \sin(\cos(\pi/2)) = \sin 0 = 0. (Composition.)

Apply the Laws ๐ŸŽฏ

Combine the Laws ๐Ÿงฎ

Given limโกxโ†’af=5\lim_{x \to a} f = 5 and limโกxโ†’ag=4\lim_{x \to a} g = 4:

1) limโกxโ†’a(f+g)=?\lim_{x \to a} (f + g) = ?

2) limโกxโ†’a(3fg)=?\lim_{x \to a} (3fg) = ?

3) limโกxโ†’af2g=?\lim_{x \to a} \dfrac{f^2}{g} = ? (decimal or fraction)

Part 4: When Direct Substitution Fails

๐Ÿšฆ When Direct Substitution Fails

Part 4 of 4 โ€” Recognize the trigger and switch tools


Topics in This Part

Section
The Indeterminate Form Trigger
What to Do Next (a Roadmap)
Substitution vs. Indeterminate vs. DNE

๐Ÿ”‘ Why this matters: When substitution gives 0/00/0 or another indeterminate form, that's a signal โ€” not an answer. You need a different technique.

โš ๏ธ The Indeterminate Form Trigger

Plugging in x=ax = a can yield:

ResultWhat it meansNext step
Clean number LLLimit is LL โœ…Done
Nonzero / 0Vertical asymptote โ€” limit DNE (or ยฑโˆž\pm\infty)Analyze sign on each side
0/00/0IndeterminateFactor, rationalize, L'Hรดpital, or known limit
โˆžโˆ’โˆž\infty - \inftyIndeterminateCombine into a single fraction first
0โ‹…โˆž0 \cdot \inftyIndeterminateRewrite as a quotient, then apply techniques
โˆž/โˆž\infty / \inftyIndeterminateCompare growth rates / divide by dominant term

๐Ÿ’ก "Indeterminate" doesn't mean "no answer." It means "more work needed."

๐Ÿ—บ๏ธ What to Do Next (A Roadmap)

Indeterminate formMost common technique
0/00/0, polynomial-onlyFactor and cancel the common factor
0/00/0, contains \sqrt{}Rationalize with the conjugate
0/00/0, contains sinโก\sin or cosโก\cosUse known special limits like limโกxโ†’0sinโกxx=1\lim_{x \to 0} \dfrac{\sin x}{x} = 1
โˆž/โˆž\infty / \inftyDivide top and bottom by highest power of xx
โˆžโˆ’โˆž\infty - \inftyCombine into a single fraction

๐Ÿ”‘ The next four micro-topics in this unit develop these techniques one at a time.

๐Ÿง  Substitution vs. Indeterminate vs. DNE

Three cleanly different outcomes when you "plug in":

  1. Number โ€” direct substitution wins, you're done.
  2. Indeterminate form โ€” you don't have the answer yet; switch to factoring/rationalizing/etc. The limit might still exist.
  3. Nonzero over zero โ€” the limit DNE because of unbounded behavior; sometimes we can describe it as ยฑโˆž\pm\infty.

โš ๏ธ Common mistake: writing "DNE" when you actually got 0/00/0. 0/00/0 is not an answer โ€” it's a request to do more work.

Diagnose the Trigger ๐ŸŽฏ

Exit Check โœ