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'Hôpital's rule, these tools form the foundation for understanding continuity, derivatives, and integrals in calculus.
Limit Laws for Function Operations
Sums and Differences
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The limit of a sum of two functions equals the sum of their limits, provided both limits exist
Mathematically: lim[f(x)+g(x)]=limf(x)+limg(x)
Example: If limf(x)=3 and limg(x)=5, then 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: lim[f(x)−g(x)]=limf(x)−limg(x)
Example: If limf(x)=7 and limg(x)=2, then 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: lim[f(x)⋅g(x)]=limf(x)⋅limg(x)
Example: If limf(x)=4 and limg(x)=6, then lim[f(x)⋅g(x)]=4⋅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: lim[f(x)/g(x)]=limf(x)/limg(x), where limg(x)=0
Example: If limf(x)=10 and limg(x)=2, then 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: lim[c⋅f(x)]=c⋅limf(x), where c is a constant
Example: If limf(x)=3 and c=4, then lim[c⋅f(x)]=4⋅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: lim[f(x)n]=[limf(x)]n, where n is a real number
Example: If limf(x)=2 and n=3, then lim[f(x)n]=23=8
Evaluating Limits of Functions
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+2x−1, then limx→2f(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)=x−2x2−4, then limx→2f(x)=limx→2x−2(x−2)(x+2)=limx→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)=3e2x, then limx→1f(x)=3e2(1)=3e2≈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)=ln(x2+1), then limx→0f(x)=ln(02+1)=ln(1)=0
When evaluating limits, be aware of potential indeterminate forms, such as 0/0, ∞/∞, 0⋅∞, ∞−∞, 00, 1∞, and ∞0, which may require further manipulation using L'Hôpital's rule or other techniques
Example: If f(x)=x−1x2−1, then limx→1f(x) is an indeterminate form of type 0/0 and can be evaluated using L'Hôpital's rule or factoring
The Squeeze Theorem for Limits
Applying the Squeeze Theorem
The Squeeze Theorem states that if f(x)≤g(x)≤h(x) for all x near a, except possibly at a, and if limf(x)=limh(x)=L as x→a, then limg(x)=L as x→a
Example: If 0≤sinx≤x for all x>0 and limx→00=limx→0x=0, then by the Squeeze Theorem, limx→0sinx=0
To apply the Squeeze Theorem, find two functions, f(x) and h(x), that "squeeze" the given function g(x) from below and above, respectively, near the point of interest
Example: To find limx→0x2cos(x1), we can use the inequality −1≤cos(x1)≤1 and squeeze x2cos(x1) between −x2 and x2
Show that the limits of the squeezing functions, f(x) and h(x), are equal as x approaches the point of interest
Example: limx→0(−x2)=limx→0x2=0
Conclude that the limit of the squeezed function, g(x), is equal to the common limit of the squeezing functions
Example: By the Squeeze Theorem, limx→0x2cos(x1)=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 limx→3x−3x2−9, factor the numerator to get limx→3x−3(x−3)(x+3), cancel the common factor, and evaluate the limit as limx→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 limx→0xx+1−1, multiply the numerator and denominator by the conjugate of the numerator, x+1+1, to get limx→0x(x+1+1)x, simplify, and evaluate the limit as limx→0x+1+11=21
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 limx→0xsin2x, use the double angle formula sin2x=2sinxcosx to get limx→02cosx, and evaluate the limit as 2
Logarithmic properties: Use the properties of logarithms to simplify expressions involving logarithmic functions before evaluating the limit
Example: To find limx→1ln(x3)lnx, use the property ln(x3)=3lnx to get limx→13lnxlnx, simplify, and evaluate the limit as 31
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 limx→0xe3x−1, let u=3x and rewrite the limit as limu→0u/3eu−1, which simplifies to 3limu→0ueu−1=3
L'Hôpital's Rule
L'Hôpital's rule: For indeterminate forms of type 0/0 or ∞/∞, differentiate the numerator and denominator separately and evaluate the limit of the resulting quotient
Example: To find limx→0xsinx, which is an indeterminate form of type 0/0, apply L'Hôpital's rule to get limx→01cosx, and evaluate the limit as 1
Example: To find limx→∞x2+1x, which is an indeterminate form of type ∞/∞, apply L'Hôpital's rule to get limx→∞2x2+12x1, simplify, and evaluate the limit as 1