Evaluating Limits Analytically (College Board AP® Calculus AB): Revision Note

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

  • $\underset{x\to a}{\lim }f\left(x\right)=f\left(a\right)$

  • 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 $f=\{\begin{array}{l}x-7,x\le 3\\ x^2,x>3\end{array}$

    • Note that $x-7$ and $x^2$ are both continuous functions on all the real numbers

    • $x=3$ is on the border for the $x-7$ piece of $f$, and 'just outside' the border for the $x^2$ piece of $f$

    • So by substituting, $\underset{x\to 3^{-}}{\lim }f\left(x\right)=\left(3\right)-7=-4$

      • and $\underset{x\to 3^{+}}{\lim }f\left(x\right)={\left(3\right)}^2=9$

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 $\frac{0}{0}$

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

  • For example by factoring and canceling common factors

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

• For example, consider $\underset{x\to 1}{\lim }\frac{\left(x-1\right)\left(x-3\right)}{x\left(x-1\right)}$

  • Substituting $x=1$ leads to $\frac{0}{0}$

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

    • We only care about its behavior near $x=1$

  • Cancel the factors

    • $\underset{x\to 1}{\lim }\frac{\left(x-1\right)\left(x-3\right)}{x\left(x-1\right)}=\underset{x\to 1}{\lim }\frac{\left(x-3\right)}{x}=-2$

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

• Not all limits for which substitution gives $\frac{0}{0}$ can be resolved this way

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

Multiplying by conjugates to find limits

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

• A quotient involving radicals can sometimes be rewritten by multiplying it top and bottom by a conjugate of either the numerator or denominator

  • The conjugate of $\sqrt{a}-\sqrt{b}$ is $\sqrt{a}+\sqrt{b}$

    • and the conjugate of $\sqrt{a}+\sqrt{b}$ is $\sqrt{a}-\sqrt{b}$

• This can be used to find limits where just using substitution gives an undefined expression like $\frac{0}{0}$

• For example, consider $\underset{x\to 0}{\lim }\frac{\sqrt{x+5}-\sqrt{5}}{x}$

  • Substituting $x=0$ leads to $\frac{0}{0}$

  • Instead, multiply the top and bottom by the conjugate of the numerator

    • $\begin{array}{rcl}\frac{\sqrt{x+5}-\sqrt{5}}{x}& =& \left(\frac{\sqrt{x+5}-\sqrt{5}}{x}\right)·\left(\frac{\sqrt{x+5}+\sqrt{5}}{\sqrt{x+5}+\sqrt{5}}\right)\end{array}$

  • Simplify

    • $\begin{array}{rcl}& & \frac{x}{x\left(\sqrt{x+5}+\sqrt{5}\right)}\end{array}$

  • Cancel

    • $\begin{array}{rcl}& & \frac{1}{\sqrt{x+5}+\sqrt{5}}\end{array}$

  • Now you can substitute $x=0$ to get $\underset{x\to 0}{\lim }\frac{\sqrt{x+5}-\sqrt{5}}{x}=\frac{1}{2\sqrt{5}}$

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

• This can be used to find a limit at infinity

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

    • $\underset{x\to \pm \infty }{\lim }\frac{k}{x^n}=0$

• For example, consider $\underset{x\to \infty }{\lim }\begin{array}{rcl}& & \frac{7x^2-4x+13}{2x^2+3x-5}\end{array}$

  • It currently looks like the limit is $\frac{\infty }{\infty }$ which is an indeterminate form

  • Multiply top and bottom by $\frac{1}{x^2}$

    • $\begin{array}{rcl}\frac{7x^2-4x+13}{2x^2+3x-5}\times \frac{\left({\displaystyle \frac{1}{x^2}}\right)}{\left({\displaystyle \frac{1}{x^2}}\right)}& =& \frac{7-{\displaystyle \frac{4}{x}}+{\displaystyle \frac{13}{x^2}}}{2+{\displaystyle \frac{3}{x}}-{\displaystyle \frac{5}{x^2}}}\end{array}$

  • Now you can take the limit of each term as $x$ tends to ∞ to get $\underset{x\to \infty }{\lim }\begin{array}{rcl}& & \frac{7x^2-4x+13}{2x^2+3x-5}\end{array}=\frac{7}{2}$