Short and Sweet Calculus

2.6 Evaluating Limits

2.6.1 Theorems for Calculating Limits

Let \(L\) and \(M\) be two finite numbers. The following relationships are true as \(x\to a\), \(x\to a^{+}\), \(x\to a^{-}\), \(x\to+\infty\), or \(x\to-\infty\)

\(\lim f(x)\) \(\lim g(x)\) \(\lim\left[f(x)+g(x)\right]\)
\(L\) \(M\) \(L+M\)
\(L\) \(+\infty\) \(+\infty\)
\(L\) \(-\infty\) \(-\infty\)
\(+\infty\) \(+\infty\) \(+\infty\)
\(-\infty\) \(-\infty\) \(-\infty\)
\(+\infty\) \(-\infty\) Indeterminate
\(\lim f(x)\) \(\lim g(x)\) \(\lim\left[f(x)g(x)\right]\)
\(L\) \(M\) \(LM\)
\(L\) \(+\infty\) \(\begin{cases} +\infty & \text{if }L>0\\ \text{\textbf{Indeterminate}} & \text{if }L=0\\ -\infty & \text{if }L<0 \end{cases}\)
\(L\) \(-\infty\) \(\begin{cases} -\infty & \text{if }L>0\\ \text{\textbf{Indeterminate}} & \text{if }L=0\\ +\infty & \text{if }L<0 \end{cases}\)
\(+\infty\) \(+\infty\) \(+\infty\)
\(-\infty\) \(+\infty\) \(-\infty\)
\(-\infty\) \(-\infty\) \(+\infty\)
\(\lim f(x)\) \(\lim g(x)\) \(\lim\dfrac{f(x)}{g(x)}\)
\(L\) \(M\neq0\) \(L/M\)
\(L\) 0 \(\begin{cases} +\infty & \text{if }Lg(x)>0\\ \text{\textbf{Indeterminate}} & \text{if }L=0\\ -\infty & \text{if }Lg(x)<0 \end{cases}\)
\(L\) \(+\infty\) or \(-\infty\) 0
\(+\infty\) or \(-\infty\) \(M\) \(\begin{cases} +\infty & \text{if }f(x)g(x)>0\\ -\infty & \text{if }f(x)g(x)<0 \end{cases}\)
\(+\infty\) or \(-\infty\) \(+\infty\) or \(-\infty\) Indeterminate

We may summarize the above theorem as:

  1. The limit of a sum is the sum of the limits. If you add a number to infinity, the result will be infinity.

  2. The limit of a product is the product of the limits. If you multiply a number (\(\neq0\)) by infinity, the result will be infinity, but if the number is negative, the sign of infinity flips. The result of zero \(\times\) infinity depends on the problem (we say this is an indeterminate limit).

  3. The limit of a quotient is the quotient of the limits provided that the limit of the denominator is not zero. If the the limit of the numerator is a number other than zero and the limit of the denominator is zero, the result will be infinity (\(+\infty\) or \(-\infty\)). If the limits of the numerator and denomiantor are both zero or are both infinity, the result may vary from one problem to another.

2.6.2 The Indeterminate Forms

Although the theorems tell us a great deal about the behavior of combined functions, as discussed above there are four cases that the theorems are silent about, specifically the cases in the above tables where there is a question mark in front of. These four cases are denoted by \[\boxed{\infty-\infty,\quad0\cdot(\pm\infty),\quad\frac{0}{0},\quad\frac{\pm\infty}{\pm\infty}}\] and are called the indeterminate forms. The value of the indeterminate forms cannot be predicted in advance. Each case may take any value (including \(+\infty\), \(-\infty\)), or may fail to exist. There are three more indeterminate forms that will be discussed in Section [sec:Ch6-L’Hopital’s-Rule].

  • In algebra 0/0 is not defined and infinity is not a number. We should emphasize that 0/0, \(0\cdot(\pm\infty)\) and so on are just shorthands for the limits shown in the above tables.

2.6.3 Other Important Theorems

2.5. (The Sandwich Theorem): If we have \[g(x)\leq f(x)\leq h(x)\] for all \(x\) in some open interval containing \(a\) except possibly at \(x=a\) itself and if \[\lim_{x\to a}g(x)=\lim_{x\to a}h(x)=L,\] then we also have \[\lim_{x\to a}f(x)=L.\]

The validity of this theorem is suggested by Figure 2.3. The rigorous proof is as follows.

The Sandwich Theorem

It follows from the Sandwich theorem that:

2.6. Suppose \(\lim_{x\to a}\alpha(x)=0\) and \(f(x)\) is a bounded function; that is, there are numbers \(m\) and \(M\) such that \(m\leq f(x)\leq m\) for \(x\) near \(a\). Then \[\lim_{x\to a}\left(\alpha(x)f(x)\right)=0.\]

For example because sine is a bounded function \[-1\leq\sin\frac{1}{x}\leq1\qquad(x\neq0)\] and \(\lim_{x\to0}x=0\), we have \[\boxed{\lim_{x\to0}x\sin\frac{1}{x}=0.}\]

2.7. If \(\lim_{x\to a}g(x)=b\) and \(f\) is continuous at the point \(b\), then \[\begin{aligned} \lim_{x\to a}f(g(x)) & =f(\lim_{x\to a}g(x))\\ & =f(b).\end{aligned}\]

2.6.4 Important Limits

  1. \({\displaystyle \lim_{x\to0}\frac{\sin x}{x}=1}\).

  2. \({\displaystyle \lim_{x\to0^{+}}\frac{1}{x}=+\infty}\) and \({\displaystyle \lim_{x\to0^{-}}\frac{1}{x}=-\infty}\).

  3. \({\displaystyle \lim_{x\to+\infty}x^{n}=+\infty}\)

  4. \({\displaystyle \lim_{x\to-\infty}x^{n}=\begin{cases} -\infty & \text{if }n\text{ is odd}\\ +\infty & \text{if }n\text{ is even} \end{cases}}\)

  5. \({\displaystyle \lim_{x\to+\infty}b^{x}=+\infty}\) (\(b>1\))

  6. \({\displaystyle \lim_{x\to-\infty}b^{x}=0}\) (\(b>1\))

  7. \({\displaystyle \lim_{x\to+\infty}b^{x}=0}\) (\(b<1\))

  8. \({\displaystyle \lim_{x\to-\infty}b^{x}=+\infty}\) (\(b<1\))

  9. \({\displaystyle \lim_{x\to0^{+}}\ln x}=-\infty\)

  10. \({\displaystyle \lim_{x\to+\infty}\ln x=+\infty}\)

2.6.5 Evaluating Limits

To evaluate the limit of an expression as \(x\to a\), we simply plug \(x=a\) into the expression, and that is the answer unless we get one of the indeterminate limits (\(0/0\), \(\pm\infty/\pm\infty\), \(0(\pm\infty\)), \(\infty-\infty\)).

If \(\lim_{x\to a}f(x)=\lim_{x\to a}g(x)=0\), to evaluate \(\lim_{x\to a}\frac{f(x)}{g(x)}\), we may need to try one or more of the following strategies:

  1. If \(f(x)\) and \(g(x)\) are both polynomials, then \((x-a)\) is a common factor. Factor out \((x-a)\) and cancel it from the denominator and numerator.1

Example 2.2. Find \[\lim_{x\to2}\frac{x^{2}-4}{3x-6}\]


Plugging \(x=2\) into the given expression \((x^{2}-4)/(3x-6)\) won’t work, because we will get 0/0. The fact that \(x^{2}-4\) and \(3x-6\) become zero upon substituting \(2\) for \(x\) shows that \(x-2\) is a common factor: \[x^{2}-4=(x-2)(x+2)\] \[3x-6=3(x-2).\] Therefore, \[\lim_{x\to2}\frac{x^{2}-4}{3x-6}=\lim_{x\to2}\frac{\cancel{(x-2)}(x+2)}{3\cancel{(x-2)}}=\frac{1}{3}\lim_{x\to2}(x+2).\] Now we can easily plug \(x=2\) into \(x+2\): \[\frac{1}{3}\lim_{x\to2}(x+2)=\frac{1}{3}(4)=\frac{4}{3}.\]

Example 2.3. Find \[\lim_{x\to3}\frac{x^{3}-27}{2x^{2}-5x-3}.\]


If we plug \(x=3\) into the given expression, we will get \(0/0\). Because \(x^{3}-27=x^{3}-3^{3}\), we can use the Difference of Cubes formula \(A^{3}-B^{3}=(A-B)(A^{2}+AB+B^{2})\) and write:2 \[x^{3}-3^{3}=(x-3)(x^{2}+3x+9).\] The numerator is a quadratic function, and we know \(x-3\) is a factor. The other factor is of the form \(ax+b\). To discover \(a\) and \(b\), we multiply it by \(x-3\) \[(x-3)(ax+b)=ax^{2}+(b-3a)x-3b.\] By comparing the above expression with \(2x^{2}-5x-3\), we realize that we must have \(a=2\) and \(b=1\). Therefore, \[2x^{2}-5x-3=(x-3)(2x+1).\] and \[\frac{x^{3}-27}{2x^{2}-5x-3}=\frac{\cancel{(x-3)}(x^{2}+3x+9)}{\cancel{(x-3)}(2x+1)}=\frac{x^{2}+3x+9}{2x+1}\qquad(\text{if }x\neq3)\] Now we can plut \(x=3\) into the simplified fraction \[\lim_{x\to3}\frac{x^{2}+3x+9}{2x+1}=\frac{3^{2}+3(3)+9}{2(3)+1}=\frac{27}{7}.\]

  1. If the denominator or numerator is an expression of the form \(\sqrt{A}-\sqrt{B}\) which becomes 0 upon substitution of \(a\) for \(x\), multiply the numerator and denominator by \(\sqrt{A}+\sqrt{B}\) and use the following formula \[\left(\sqrt{A}-\sqrt{B}\right)\left(\sqrt{A}+\sqrt{B}\right)=A-B.\]

Example 2.4. Find \[\lim_{x\to-4}\frac{x^{2}-16}{3-\sqrt{x^{2}-7}}.\]


If we plug \(x=-4\), we will get the indeterminate form \(0/0\). As discussed above, we multiply and divide the given expression by the conjugat of the denominator, which is \(3+\sqrt{x^{2}-7}\): \[\begin{aligned} \lim_{x\to-4}\frac{x^{2}-16}{3-\sqrt{x^{2}-7}} & =\lim_{x\to-4}\left(\frac{x^{2}-16}{3-\sqrt{x^{2}-7}}\cdot\frac{3+\sqrt{x^{2}-7}}{3+\sqrt{x^{2}-7}}\right)=\lim_{x\to-4}\frac{(x^{2}-16)\left(3+\sqrt{x^{2}-7}\right)}{3^{2}-(x^{2}-7)}\\ & =\lim_{x\to-4}\frac{(x^{2}-16)\left(3+\sqrt{x^{2}-7}\right)}{16-x^{2}}=\lim_{x\to-4}\left(-3-\sqrt{x^{2}-7}\right)=-3-3=-6.\end{aligned}\]

  1. If it is not easy to factor out \((x-a)\), you may use the long division and divide \(f(x)\) and \(g(x)\) by \(x-a\).↩︎

  2. If we could not remember the Difference of Cubes formula, we could also divide the numerator by \((x-3)\):


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