Evaluating limits

Indeterminate forms

These are forms the value of which cannot be determined unless we know the exact context in which they occur. There are seven indeterminate forms \[\dfrac{0}{0};\quad \dfrac{\infty}{\infty};\quad \infty-\infty;\quad 0\cdot\infty;\quad \infty^0;\quad 0^0;\quad 1^\infty\]

Note that in the context of limits \(0\) means an infinitesimal. It does not mean exactly zero. Also, \(1\) means a value approaching one. It does not mean exactly one.

Reciprocals of infinites and infinitesmals

\[\dfrac{1}{+\infty}=0^+;\qquad\dfrac{1}{-\infty}=0^-;\qquad\dfrac{1}{0^+}=+\infty;\qquad\dfrac{1}{0^-}=-\infty\]

Multplying by \(-1\)

The result is simply flipping the sign of the other multiplicand. \[-1\times 0^+=0^-;\qquad -1\times 0^-=0^+;\qquad -1\times +\infty=-\infty;\qquad -1\times -\infty=+\infty\]

Adding infinities

\[+\infty+\infty=+\infty;\qquad\qquad -\infty-\infty=-\infty\]

Adding infinitesimals

The result of adding a finite value \(x\in\mathbb{R}\) to a positive infinitesimal is approaching this value from above, and to a negative infinitesimal is approaching this value from below. \[x+0^+=x^+;\qquad\qquad x+0^-=x^-\]

The result of adding an infinitesimal to \(\pm\infty\) is \(\pm\infty\) \[\pm\infty+0^+=\pm\infty;\qquad\qquad \pm\infty+0^-=\pm\infty\]

The result of adding an infinitesimal to another infinitesimal is an infinitesimal. The result is positive if both infinitesimals are positive, and negative if both are negative. If the added infinitesimals have different signs, the result is that of the larger infinitesimal. \[0^++0^+=0^+;\qquad 0^-+0^-=0^-;\qquad 0^++0^-=\begin{cases} 0^+;& 0^+\gt 0^- \\ 0^-;& 0^+\lt 0^- \end{cases}\]

Multiplying by infinity

Multiplying a non-zero, real number \(x\in\mathbb{R}\) by infinity results in infinity. Multiplying infinity by infinity results in infinity. The sign of the result depends on the sign of the multiplicands. \[x\times\pm\infty=\begin{cases} \pm\infty;& x\gt 0 \\ \mp\infty;& x\lt 0 \end{cases}\] \[\pm\infty\times\pm\infty=+\infty;\qquad\qquad \pm\infty\times\mp\infty=-\infty\]

Multiplying by an infinitesimal

Multiplying a non-zero, real number \(x\in\mathbb{R}\) by an infinitesimal results in an infinitesimal. The sign of the result depends on the signs of the multiplicands. \[x\times 0^+=\begin{cases} 0^+;& x\gt 0 \\ 0^-;& x\lt 0 \end{cases}\qquad\qquad x\times 0^-=\begin{cases} 0^-;& x\gt 0 \\ 0^+;& x\lt 0 \end{cases}\]

Multiplying two infinitesimals results in another infinitesimal. The sign of the result depends on the signs of the multiplicands. \[0^+\times0^+=0^+;\qquad 0^+\times0^-=0^-;\qquad 0^-\times0^+=0^-;\qquad 0^-\times0^-=0^+\]

Raising to infinitesimal power

As long as the base is a non-zero, finite (not infinitesimal) value, the result approaches \(1\). The base must be a non-unit, positive number by definition. \[b^{0^+}=\begin{cases} 1^+;& b\gt 1 \\ 1^-;& 0\lt b\lt 1 \end{cases}\qquad\qquad b^{0^-}=\begin{cases} 1^-;& b\gt 1 \\ 1^+;& 0\lt b\lt 1 \end{cases}\]

Direct substitution

As long as the function is defined at the limit point \(a\) in \(\lim\limits_{x\to a}f(x)\), you should try direct substitution first. If \(a\) is the boundary of an interval in the definition of a piecewise function, the one-sided limits may not be the same. If substitution results in an indeterminate form, you should proceed with the following steps.

Factoring

This is typically considered when we get the indeterminate form \(\frac{0}{0}\). Factoring may help eliminate the factor that results in the zero in both the numerator and the denominator. For example, \begin{align} \lim\limits_{x\to1}\dfrac{x-1}{x^2-1}&=\lim\limits_{x\to1}\dfrac{x-1}{(x+1)(x-1)} \\ &=\lim\limits_{x\to1}\dfrac{1}{x+1}=\dfrac{1}{1+1}=\dfrac{1}{2} \end{align}

Rational functions as \(x\to\pm\infty\)

Typically, the easiest way is to divide both the numerator and the denominator by the highest power of \(x\). This makes all terms evaluate to zero except those with the highest power of \(x\). For example, \begin{align} \lim\limits_{x\to\infty}\dfrac{2x^2+x}{x^2-1} &= \lim\limits_{x\to\infty}\dfrac{\frac{2x^2}{x^2}+\frac{x}{x^2}}{\frac{x^2}{x^2}-\frac{1}{x^2}} \\ \lim\limits_{x\to\infty}\dfrac{2+\frac{1}{x}}{1-\frac{1}{x^2}}=\dfrac{2+0}{1-0}=2 \end{align}

Multiplying by the conjugate

This is typically considered when we have square roots in the numerator or the denominator. The conjugate of \((a\pm b)\) is \((a\mp b)\) (with the sign between the two terms flipped). Multiplying a binomial by its conjugate gives a difference of squares. When one of the terms in the binomial is a square root, the root is eliminated. For example, \begin{align} \lim\limits_{x\to4}\dfrac{x-4}{\sqrt{x}-2} &= \lim\limits_{x\to4}\dfrac{x-4}{\sqrt{x}-2}\cdot\dfrac{\sqrt{x}+2}{\sqrt{x}+2} \\ &=\lim\limits_{x\to4}\dfrac{(x-4)(\sqrt{x}+2)}{x-4} \\ &=\lim\limits_{x\to4}(\sqrt{x}+2)=\sqrt{4}+2=4 \end{align}

Change of variable

This is helpful in many situations where we have roots higher than the square root. Remember to change the limit point also, not just the expression of the function. For example, consider \[\lim\limits_{x\to 8}\dfrac{x-8}{\sqrt[3]{x}-2}\] which evaluates to an indeterminate form. Let us do the change-of-variable \(x=y^3\). This will eliminate the cube root. As \(x\to8\), \(y^3\to8\), which means \(y\to 2\). Therefore, the limit becomes \begin{align}\require{cancel} \lim\limits_{y\to2}\dfrac{y^3-8}{y-2}&=\lim\limits_{y\to2}\dfrac{\cancel{(y-2)}(y^2+2y+4)}{\cancel{y-2}} \\ &=\lim\limits_{y\to2}(y^2+2y+4)=2^2+2(2)+4=12 \end{align}

This method may also be helpful if we have similar brackets raised to powers and we want to avoid expanding the brackets. Consider \[\lim\limits_{x\to 1}\dfrac{(x^2-3x+4)^4-16}{(x^2-3x+4)^3-8}\] which evaluates to an indeterminate form. Now let \(y=x^2-3x+4\). As \(x\to 1\), we can see that \(y\to 2\). The limit then becomes \begin{align} \lim\limits_{x\to1}\dfrac{(x^2-3x+4)^4-16}{(x^2-3x+4)^3-8}&=\lim\limits_{y\to 2}\dfrac{y^4-16}{y^3-8} \\ &=\lim\limits_{y\to 2}\dfrac{(y^2-4)(y^2+4)}{(y-2)(y^2+2y+4)} \\ &=\lim\limits_{y\to 2}\dfrac{\cancel{(y-2)}(y+2)(y^2+4)}{\cancel{(y-2)}(y^2+2y+4)} \\ &=\lim\limits_{y\to 2}\dfrac{(y+2)(y^2+4)}{y^2+2y+4} \\ &= \dfrac{(2+2)(2^2+4)}{2^2+2(2)+4}=\dfrac{(4)(8)}{12}=\dfrac{8}{3} \end{align}