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๐Ÿƒ๐Ÿฝโ€โ™€๏ธโ€โžก๏ธIntro to Mathematical Analysis Unit 5 Review

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5.3 Limit Theorems for Functions

5.3 Limit Theorems for Functions

Written by the Fiveable Content Team โ€ข Last updated August 2025
Written by the Fiveable Content Team โ€ข Last updated August 2025
๐Ÿƒ๐Ÿฝโ€โ™€๏ธโ€โžก๏ธIntro to Mathematical Analysis
Unit & Topic Study Guides

Limit theorems for functions are crucial tools in calculus. They help us understand how functions behave as we approach specific points or infinity. These rules simplify complex limit calculations by breaking them down into smaller, more manageable parts.

By mastering these theorems, you'll be able to tackle a wide range of limit problems. From basic arithmetic operations to more advanced techniques like L'Hรดpital's rule, these tools form the foundation for understanding continuity, derivatives, and integrals in calculus.

Limit Laws for Function Operations

Sums and Differences

  • The limit of a sum of two functions equals the sum of their limits, provided both limits exist
    • Mathematically: limโก[f(x)+g(x)]=limโกf(x)+limโกg(x)\lim[f(x) + g(x)] = \lim f(x) + \lim g(x)
    • Example: If limโกf(x)=3\lim f(x) = 3 and limโกg(x)=5\lim g(x) = 5, then limโก[f(x)+g(x)]=3+5=8\lim[f(x) + g(x)] = 3 + 5 = 8
  • The limit of a difference of two functions equals the difference of their limits, provided both limits exist
    • Mathematically: limโก[f(x)โˆ’g(x)]=limโกf(x)โˆ’limโกg(x)\lim[f(x) - g(x)] = \lim f(x) - \lim g(x)
    • Example: If limโกf(x)=7\lim f(x) = 7 and limโกg(x)=2\lim g(x) = 2, then limโก[f(x)โˆ’g(x)]=7โˆ’2=5\lim[f(x) - g(x)] = 7 - 2 = 5

Products, Quotients, and Powers

  • The limit of a product of two functions equals the product of their limits, provided both limits exist
    • Mathematically: limโก[f(x)โ‹…g(x)]=limโกf(x)โ‹…limโกg(x)\lim[f(x) \cdot g(x)] = \lim f(x) \cdot \lim g(x)
    • Example: If limโกf(x)=4\lim f(x) = 4 and limโกg(x)=6\lim g(x) = 6, then limโก[f(x)โ‹…g(x)]=4โ‹…6=24\lim[f(x) \cdot g(x)] = 4 \cdot 6 = 24
  • The limit of a quotient of two functions equals the quotient of their limits, provided the limit of the denominator is non-zero
    • Mathematically: limโก[f(x)/g(x)]=limโกf(x)/limโกg(x)\lim[f(x) / g(x)] = \lim f(x) / \lim g(x), where limโกg(x)โ‰ 0\lim g(x) \neq 0
    • Example: If limโกf(x)=10\lim f(x) = 10 and limโกg(x)=2\lim g(x) = 2, then limโก[f(x)/g(x)]=10/2=5\lim[f(x) / g(x)] = 10 / 2 = 5
  • The limit of a constant multiple of a function equals the constant multiple of the limit of the function
    • Mathematically: limโก[cโ‹…f(x)]=cโ‹…limโกf(x)\lim[c \cdot f(x)] = c \cdot \lim f(x), where cc is a constant
    • Example: If limโกf(x)=3\lim f(x) = 3 and c=4c = 4, then limโก[cโ‹…f(x)]=4โ‹…3=12\lim[c \cdot f(x)] = 4 \cdot 3 = 12
  • The limit of a function raised to a power equals the limit of the function raised to that power, provided the limit exists
    • Mathematically: limโก[f(x)n]=[limโกf(x)]n\lim[f(x)^n] = [\lim f(x)]^n, where nn is a real number
    • Example: If limโกf(x)=2\lim f(x) = 2 and n=3n = 3, then limโก[f(x)n]=23=8\lim[f(x)^n] = 2^3 = 8

Evaluating Limits of Functions

Sums and Differences, Sum and Difference Identities โ€“ Algebra and Trigonometry OpenStax

Polynomial, Rational, and Exponential Functions

  • To find the limit of a polynomial function, evaluate the function at the point of interest by substituting the limiting value for the variable
    • Example: If f(x)=3x2+2xโˆ’1f(x) = 3x^2 + 2x - 1, then limโกxโ†’2f(x)=3(2)2+2(2)โˆ’1=15\lim_{x \to 2} f(x) = 3(2)^2 + 2(2) - 1 = 15
  • To find the limit of a rational function, factor and cancel common factors in the numerator and denominator, then evaluate the simplified function at the point of interest
    • Example: If f(x)=x2โˆ’4xโˆ’2f(x) = \frac{x^2 - 4}{x - 2}, then limโกxโ†’2f(x)=limโกxโ†’2(xโˆ’2)(x+2)xโˆ’2=limโกxโ†’2(x+2)=4\lim_{x \to 2} f(x) = \lim_{x \to 2} \frac{(x - 2)(x + 2)}{x - 2} = \lim_{x \to 2} (x + 2) = 4
  • To find the limit of an exponential function, use the properties of exponents and evaluate the function at the point of interest
    • Example: If f(x)=3e2xf(x) = 3e^{2x}, then limโกxโ†’1f(x)=3e2(1)=3e2โ‰ˆ22.17\lim_{x \to 1} f(x) = 3e^{2(1)} = 3e^2 \approx 22.17

Logarithmic Functions and Indeterminate Forms

  • To find the limit of a logarithmic function, use the properties of logarithms and evaluate the function at the point of interest
    • Example: If f(x)=lnโก(x2+1)f(x) = \ln(x^2 + 1), then limโกxโ†’0f(x)=lnโก(02+1)=lnโก(1)=0\lim_{x \to 0} f(x) = \ln(0^2 + 1) = \ln(1) = 0
  • When evaluating limits, be aware of potential indeterminate forms, such as 0/00/0, โˆž/โˆž\infty/\infty, 0โ‹…โˆž0 \cdot \infty, โˆžโˆ’โˆž\infty - \infty, 000^0, 1โˆž1^\infty, and โˆž0\infty^0, which may require further manipulation using L'Hรดpital's rule or other techniques
    • Example: If f(x)=x2โˆ’1xโˆ’1f(x) = \frac{x^2 - 1}{x - 1}, then limโกxโ†’1f(x)\lim_{x \to 1} f(x) is an indeterminate form of type 0/00/0 and can be evaluated using L'Hรดpital's rule or factoring

The Squeeze Theorem for Limits

Sums and Differences, I was wondering why the statement in the title is true only if the functions we are dealing with ...

Applying the Squeeze Theorem

  • The Squeeze Theorem states that if f(x)โ‰คg(x)โ‰คh(x)f(x) \leq g(x) \leq h(x) for all xx near aa, except possibly at aa, and if limโกf(x)=limโกh(x)=L\lim f(x) = \lim h(x) = L as xโ†’ax \to a, then limโกg(x)=L\lim g(x) = L as xโ†’ax \to a
    • Example: If 0โ‰คsinโกxโ‰คx0 \leq \sin x \leq x for all x>0x > 0 and limโกxโ†’00=limโกxโ†’0x=0\lim_{x \to 0} 0 = \lim_{x \to 0} x = 0, then by the Squeeze Theorem, limโกxโ†’0sinโกx=0\lim_{x \to 0} \sin x = 0
  • To apply the Squeeze Theorem, find two functions, f(x)f(x) and h(x)h(x), that "squeeze" the given function g(x)g(x) from below and above, respectively, near the point of interest
    • Example: To find limโกxโ†’0x2cosโก(1x)\lim_{x \to 0} x^2 \cos(\frac{1}{x}), we can use the inequality โˆ’1โ‰คcosโก(1x)โ‰ค1-1 \leq \cos(\frac{1}{x}) \leq 1 and squeeze x2cosโก(1x)x^2 \cos(\frac{1}{x}) between โˆ’x2-x^2 and x2x^2
  • Show that the limits of the squeezing functions, f(x)f(x) and h(x)h(x), are equal as xx approaches the point of interest
    • Example: limโกxโ†’0(โˆ’x2)=limโกxโ†’0x2=0\lim_{x \to 0} (-x^2) = \lim_{x \to 0} x^2 = 0
  • Conclude that the limit of the squeezed function, g(x)g(x), is equal to the common limit of the squeezing functions
    • Example: By the Squeeze Theorem, limโกxโ†’0x2cosโก(1x)=0\lim_{x \to 0} x^2 \cos(\frac{1}{x}) = 0

Limit Problems with Algebraic Techniques

Factoring and Rationalization

  • Factoring: Factor the numerator and denominator of a rational function to cancel common factors and simplify the expression before evaluating the limit
    • Example: To find limโกxโ†’3x2โˆ’9xโˆ’3\lim_{x \to 3} \frac{x^2 - 9}{x - 3}, factor the numerator to get limโกxโ†’3(xโˆ’3)(x+3)xโˆ’3\lim_{x \to 3} \frac{(x - 3)(x + 3)}{x - 3}, cancel the common factor, and evaluate the limit as limโกxโ†’3(x+3)=6\lim_{x \to 3} (x + 3) = 6
  • Rationalization: Multiply the numerator and denominator by the conjugate of the denominator to rationalize the denominator and simplify the expression before evaluating the limit
    • Example: To find limโกxโ†’0x+1โˆ’1x\lim_{x \to 0} \frac{\sqrt{x + 1} - 1}{x}, multiply the numerator and denominator by the conjugate of the numerator, x+1+1\sqrt{x + 1} + 1, to get limโกxโ†’0xx(x+1+1)\lim_{x \to 0} \frac{x}{x(\sqrt{x + 1} + 1)}, simplify, and evaluate the limit as limโกxโ†’01x+1+1=12\lim_{x \to 0} \frac{1}{\sqrt{x + 1} + 1} = \frac{1}{2}

Trigonometric Identities, Logarithmic Properties, and Change of Variable

  • Trigonometric identities: Use trigonometric identities to simplify expressions involving trigonometric functions before evaluating the limit
    • Example: To find limโกxโ†’0sinโก2xx\lim_{x \to 0} \frac{\sin 2x}{x}, use the double angle formula sinโก2x=2sinโกxcosโกx\sin 2x = 2\sin x \cos x to get limโกxโ†’02cosโกx\lim_{x \to 0} 2\cos x, and evaluate the limit as 22
  • Logarithmic properties: Use the properties of logarithms to simplify expressions involving logarithmic functions before evaluating the limit
    • Example: To find limโกxโ†’1lnโกxlnโก(x3)\lim_{x \to 1} \frac{\ln x}{\ln(x^3)}, use the property lnโก(x3)=3lnโกx\ln(x^3) = 3\ln x to get limโกxโ†’1lnโกx3lnโกx\lim_{x \to 1} \frac{\ln x}{3\ln x}, simplify, and evaluate the limit as 13\frac{1}{3}
  • Change of variable: Introduce a new variable to simplify the expression or to make the limit more apparent before evaluating the limit
    • Example: To find limโกxโ†’0e3xโˆ’1x\lim_{x \to 0} \frac{e^{3x} - 1}{x}, let u=3xu = 3x and rewrite the limit as limโกuโ†’0euโˆ’1u/3\lim_{u \to 0} \frac{e^u - 1}{u/3}, which simplifies to 3limโกuโ†’0euโˆ’1u=33\lim_{u \to 0} \frac{e^u - 1}{u} = 3

L'Hรดpital's Rule

  • L'Hรดpital's rule: For indeterminate forms of type 0/00/0 or โˆž/โˆž\infty/\infty, differentiate the numerator and denominator separately and evaluate the limit of the resulting quotient
    • Example: To find limโกxโ†’0sinโกxx\lim_{x \to 0} \frac{\sin x}{x}, which is an indeterminate form of type 0/00/0, apply L'Hรดpital's rule to get limโกxโ†’0cosโกx1\lim_{x \to 0} \frac{\cos x}{1}, and evaluate the limit as 11
    • Example: To find limโกxโ†’โˆžxx2+1\lim_{x \to \infty} \frac{x}{\sqrt{x^2 + 1}}, which is an indeterminate form of type โˆž/โˆž\infty/\infty, apply L'Hรดpital's rule to get limโกxโ†’โˆž12x2x2+1\lim_{x \to \infty} \frac{1}{\frac{2x}{2\sqrt{x^2 + 1}}}, simplify, and evaluate the limit as 11