AP®MathsCollege BoardCalculus ABRevision NotesUnit 1: Limits & ContinuityLimitsEvaluating Limits Analytically

Evaluating Limits Analytically (College Board AP® Calculus AB)

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Roger B

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Roger B

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Substituting to find limits

How can I use substitution to find limits?

  • If a function f is continuous on an open interval, and if a is a number contained in that interval, then

    • limit as x rightwards arrow a of f open parentheses x close parentheses equals f open parentheses a close parentheses

    • I.e. for a continuous function, the limit at a point is equal to the value of the function at that point

  • This idea can be extended to a point that is on or 'just outside' the border of an interval on which a function (or part of a function) is defined

    • This is especially useful for one-sided limits at the 'joins' of piecewise functions

    • E.g. the function f defined by f equals open curly brackets table row cell x minus 7 comma space space x less or equal than 3 end cell row cell x squared comma space space x greater than 3 end cell end table close

      • Note that x minus 7 and x squared are both continuous functions on all the real numbers

      • x equals 3 is on the border for the x minus 7 piece of f, and 'just outside' the border for the x squared piece of f

      • So by substituting, limit as x rightwards arrow 3 to the power of minus of f open parentheses x close parentheses equals open parentheses 3 close parentheses minus 7 equals negative 4

        • and limit as x rightwards arrow 3 to the power of plus of f open parentheses x close parentheses equals open parentheses 3 close parentheses squared equals 9

Worked Example

Let f be the function defined by f open parentheses x close parentheses equals open curly brackets table row cell x cubed minus 3 x plus 4 comma space space x less than negative 2 end cell row cell x squared minus 2 comma space space x greater or equal than negative 2 end cell end table close

Find:

(a) limit as x rightwards arrow 4 of f open parentheses x close parentheses

Answer:

x equals 4 belongs to the x squared minus 2 piece of the function, and x squared minus 2 is continuous on the open interval open parentheses negative 2 comma space infinity close parentheses

So the limit is just equal to the value of the function at x equals 4

limit as x rightwards arrow 4 of f open parentheses x close parentheses equals f open parentheses 4 close parentheses equals open parentheses 4 close parentheses squared minus 2

limit as x rightwards arrow 4 of f open parentheses x close parentheses equals 14

(b) limit as x rightwards arrow negative 2 of f open parentheses x close parentheses

For this limit to exist, the limits from the left and right must both exist and be equal

x equals negative 2 is 'just outside' the border for the x cubed minus 3 x plus 4 piece of f, and on the border for the x squared minus 2 piece of f

So use substitution to find the one-sided limits

limit as x rightwards arrow negative 2 to the power of minus of f open parentheses x close parentheses equals open parentheses negative 2 close parentheses cubed minus 3 open parentheses negative 2 close parentheses plus 4 equals negative 8 plus 6 plus 4 equals 2 limit as x rightwards arrow negative 2 to the power of plus of f open parentheses x close parentheses equals open parentheses negative 2 close parentheses squared minus 2 equals 4 minus 2 equals 2

Those limits are equal, so that is the value of the (two-sided) limit at x equals negative 2

limit as x rightwards arrow negative 2 of f open parentheses x close parentheses equals 2

Simplifying to find limits

How can I simplify functions to find limits?

  • Sometimes trying to evaluate a limit by substitution gives an answer that is not defined

    • E.g. returning a value like 0 over 0

  • In these cases, sometimes the equation of the function can be simplified

    • For example by factorising and cancelling common factors

    • After this substitution may give a well-defined value for the limit

  • E.g. let f be the function defined by f open parentheses x close parentheses equals fraction numerator open parentheses x minus 1 close parentheses open parentheses x minus 3 close parentheses over denominator x open parentheses x minus 1 close parentheses end fraction

    • This function is undefined at x equals 1, because f open parentheses 1 close parentheses equals 0 over 0

    • But to find the limit at x equals 1 we don't care about the function's value at x equals 1

      • We only care about it's behavior near x equals 1

    • So we can cancel the common factor open parentheses x minus 1 close parentheses and proceed like this:

      • limit as x rightwards arrow 1 of f open parentheses x close parentheses equals right enclose fraction numerator open parentheses x minus 1 close parentheses open parentheses x minus 3 close parentheses over denominator x open parentheses x minus 1 close parentheses end fraction end enclose subscript x equals 1 end subscript equals right enclose fraction numerator open parentheses x minus 3 close parentheses over denominator x end fraction end enclose subscript x equals 1 end subscript equals fraction numerator open parentheses 1 close parentheses minus 3 over denominator open parentheses 1 close parentheses end fraction equals negative 2

        • The 'vertical bar notation' right enclose blank end enclose subscript x equals 1 end subscript means to evaluate the expression to the left of the bar at x equals 1

    • Note that the limit can be well-defined even at a point where the function itself is undefined!

    • Also note that f open parentheses x close parentheses equals fraction numerator open parentheses x minus 1 close parentheses open parentheses x minus 3 close parentheses over denominator x open parentheses x minus 1 close parentheses end fraction and g open parentheses x close parentheses equals fraction numerator open parentheses x minus 3 close parentheses over denominator x end fraction are not the same function

      • The function g is defined at x equals 1, and the function f isn't

      • But for purposes of calculating limits, canceling common factors to make f 'look like' g is perfectly okay

  • Not all limits for which substitution gives 0 over 0 can be resolved this way

    • See the L'Hospital's Rule study guide for another method to deal with such limits

Worked Example

Let f be the function defined by f open parentheses x close parentheses equals fraction numerator x squared plus 2 x over denominator x squared minus x minus 6 end fraction.

Find limit as x rightwards arrow negative 2 of f open parentheses x close parentheses.

Answer:

Attempting to evaluate this limit by substitution doesn't work: fraction numerator open parentheses negative 2 close parentheses squared plus 2 open parentheses negative 2 close parentheses over denominator open parentheses negative 2 close parentheses squared minus open parentheses negative 2 close parentheses minus 6 end fraction equals fraction numerator 4 minus 4 over denominator 4 plus 2 minus 6 end fraction equals 0 over 0

Instead, factorise the numerator and denominator, simplify the fraction, and use substitution to evaluate the limit in the simplified form

fraction numerator x squared plus 2 x over denominator x squared minus x minus 6 end fraction equals fraction numerator x open parentheses x plus 2 close parentheses over denominator open parentheses x plus 2 close parentheses open parentheses x minus 3 close parentheses end fraction equals fraction numerator x over denominator x minus 3 end fraction fraction numerator open parentheses negative 2 close parentheses over denominator open parentheses negative 2 close parentheses minus 3 end fraction equals fraction numerator negative 2 over denominator negative 5 end fraction equals 2 over 5

limit as x rightwards arrow negative 2 of f open parentheses x close parentheses equals 2 over 5

Multiplying by conjugates to find limits

How can I use multiplying by conjugates to help find limits?

  • A quotient involving surds (i.e. square roots) can sometimes be rewritten by multiplying it top and bottom by a conjugate of either the numerator or denominator

    • The conjugate of square root of a minus square root of b is square root of a plus square root of b

      • and the conjugate of square root of a plus square root of b is square root of a minus square root of b

  • For example,

    table row cell fraction numerator square root of 5 minus square root of x plus 5 end root over denominator x end fraction end cell equals cell fraction numerator square root of x plus 5 end root minus square root of 5 over denominator x end fraction open parentheses fraction numerator square root of x plus 5 end root plus square root of 5 over denominator square root of x plus 5 end root plus square root of 5 end fraction close parentheses end cell row blank equals cell fraction numerator x over denominator x open parentheses square root of x plus 5 end root plus square root of x close parentheses end fraction end cell row blank equals cell fraction numerator 1 over denominator square root of x plus 5 end root plus square root of x end fraction end cell end table

  • This can be used to find limits where just using substitution gives an undefined expression like 0 over 0

Worked Example

Let f be the function defined by f equals fraction numerator x over denominator square root of 3 minus square root of x plus 3 end root end fraction comma space space x greater or equal than negative 3.

Find limit as x rightwards arrow 0 of f open parentheses x close parentheses.

Answer:

Attempting to evaluate this limit by substitution doesn't work: fraction numerator open parentheses 0 close parentheses over denominator square root of 3 minus square root of open parentheses 0 close parentheses plus 3 end root end fraction equals fraction numerator 0 over denominator square root of 3 minus square root of 3 end fraction equals 0 over 0

Instead, multiply the top and bottom by the conjugate of the denominator, then simplify the fraction

table row cell fraction numerator x over denominator square root of 3 minus square root of x plus 3 end root end fraction end cell equals cell fraction numerator x over denominator square root of 3 minus square root of x plus 3 end root end fraction open parentheses fraction numerator square root of 3 plus square root of x plus 3 end root over denominator square root of 3 plus square root of x plus 3 end root end fraction close parentheses end cell row blank equals cell fraction numerator x open parentheses square root of 3 plus square root of x plus 3 end root close parentheses over denominator 3 minus open parentheses x plus 3 close parentheses end fraction end cell row blank equals cell fraction numerator x open parentheses square root of 3 plus square root of x plus 3 end root close parentheses over denominator negative x end fraction end cell row blank equals cell negative square root of 3 minus square root of x plus 3 end root end cell end table

Now use substitution to evaluate the limit in the simplified form

table row cell negative square root of 3 minus square root of open parentheses 0 close parentheses plus 3 end root end cell equals cell negative square root of 3 minus square root of 3 equals negative 2 square root of 3 end cell end table

limit as x rightwards arrow 0 of f open parentheses x close parentheses equals negative 2 square root of 3

Multiplying by reciprocals to find limits

How can I use multiplying by reciprocals to find a limit at infinity?

  • A quotient involving powers of x can be rewritten by multiplying its top and bottom by the reciprocal of the highest power of x in the numerator or denominator

  • For example,

    table row cell space space space space space space space space space fraction numerator 7 x squared minus 4 x plus 13 over denominator 2 x squared plus 3 x minus 5 end fraction end cell equals cell fraction numerator 7 x squared minus 4 x plus 13 over denominator 2 x squared plus 3 x minus 5 end fraction cross times fraction numerator open parentheses 1 over x squared close parentheses over denominator open parentheses 1 over x squared close parentheses end fraction end cell row blank equals cell fraction numerator 7 minus 4 over x plus 13 over x squared over denominator 2 plus 3 over x minus 5 over x squared end fraction end cell end table

  • This can be used to find a limit at infinity

    • because if k is any constant and n is a positive integer, then

      • limit as x rightwards arrow plus-or-minus infinity of k over x to the power of n equals 0

Worked Example

Let f be the function defined by f open parentheses x close parentheses equals fraction numerator 4 x cubed minus 3 x squared plus 1 over denominator 5 x cubed plus 2 x plus 17 end fraction.

Find limit as x rightwards arrow infinity of f open parentheses x close parentheses.

Answer:

The numerator and denominator of the fraction both tend to infinity as x tends to infinity, but infinity over infinity is not defined as a limit (see the study guide on L'Hospital's Rule for another way to deal with limits like this)

Instead, multiply the top and bottom by the reciprocal of the highest power, x cubed

table row cell fraction numerator 4 x cubed minus 3 x squared plus 1 over denominator 5 x cubed plus 2 x plus 17 end fraction end cell equals cell fraction numerator 4 x cubed minus 3 x squared plus 1 over denominator 5 x cubed plus 2 x plus 17 end fraction cross times fraction numerator open parentheses 1 over x cubed close parentheses over denominator open parentheses 1 over x cubed close parentheses end fraction end cell row blank equals cell fraction numerator 4 minus 3 over x plus 1 over x cubed over denominator 5 plus 2 over x squared plus 17 over x cubed end fraction end cell end table

Now take the limit of that rewritten form

All the terms in the numerator and denominator tend towards zero as x tends towards infinity, except for the constant terms (4 in the numerator and 5 in the denominator)

limit as x rightwards arrow infinity of open parentheses table row blank blank cell fraction numerator 4 minus 3 over x plus 1 over x cubed over denominator 5 plus 2 over x squared plus 17 over x cubed end fraction end cell end table close parentheses equals fraction numerator 4 minus 0 plus 0 over denominator 5 plus 0 plus 0 end fraction

limit as x rightwards arrow infinity of f open parentheses x close parentheses equals 4 over 5

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    Roger B

    Author: Roger B

    Roger's teaching experience stretches all the way back to 1992, and in that time he has taught students at all levels between Year 7 and university undergraduate. Having conducted and published postgraduate research into the mathematical theory behind quantum computing, he is more than confident in dealing with mathematics at any level the exam boards might throw at you.