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Section 2.4 : Limit Properties

2. Given \(\mathop {\lim }\limits_{x \to - 4} f\left( x \right) = 1\), \(\mathop {\lim }\limits_{x \to - 4} g\left( x \right) = 10\) and \(\mathop {\lim }\limits_{x \to - 4} h\left( x \right) = - 7\) use the limit properties given in this section to compute each of the following limits. If it is not possible to compute any of the limits clearly explain why not.

  1. \(\displaystyle \mathop {\lim }\limits_{x \to - 4} \left[ {\frac{{f\left( x \right)}}{{g\left( x \right)}} - \frac{{h\left( x \right)}}{{f\left( x \right)}}} \right]\)
  2. \(\mathop {\lim }\limits_{x \to - 4} \left[ {f\left( x \right)g\left( x \right)h\left( x \right)} \right]\)
  3. \(\displaystyle \mathop {\lim }\limits_{x \to - 4} \left[ {\frac{1}{{h\left( x \right)}} + \frac{{3 - f\left( x \right)}}{{g\left( x \right) + h\left( x \right)}}} \right]\)
  4. \(\displaystyle \mathop {\lim }\limits_{x \to - 4} \left[ {2h\left( x \right) - \frac{1}{{h\left( x \right) + 7f\left( x \right)}}} \right]\)
Hint : For each of these all we need to do is use the limit properties on the limit until the given limits appear and we can then compute the value.

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a \(\displaystyle \mathop {\lim }\limits_{x \to - 4} \left[ {\frac{{f\left( x \right)}}{{g\left( x \right)}} - \frac{{h\left( x \right)}}{{f\left( x \right)}}} \right]\)Show Solution

Here is the work for this limit. At each step the property (or properties) used are listed and note that in some cases the properties may have been used more than once in the indicated step.

\[\begin{alignat*}{3}\mathop {\lim }\limits_{x \to - 4} \left[ {\frac{{f\left( x \right)}}{{g\left( x \right)}} - \frac{{h\left( x \right)}}{{f\left( x \right)}}} \right] & = \mathop {\lim }\limits_{x \to - 4} \frac{{f\left( x \right)}}{{g\left( x \right)}} - \mathop {\lim }\limits_{x \to - 4} \frac{{h\left( x \right)}}{{f\left( x \right)}} & & \hspace{0.25in}{\mbox{Property 2}}\\ & = \frac{{\mathop {\lim }\limits_{x \to - 4} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to - 4} g\left( x \right)}} - \frac{{\mathop {\lim }\limits_{x \to - 4} h\left( x \right)}}{{\mathop {\lim }\limits_{x \to - 4} f\left( x \right)}} & & \hspace{0.25in}{\mbox{Property 4}}\\ & = \frac{1}{{10}} - \frac{{ - 7}}{1} & & \hspace{0.25in}{\mbox{Plug in values of limits}}\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{71}}{{10}}}} & & & \end{alignat*}\]

Note that were able to use Property 4 in the second step only because after we evaluated the limit of the denominators (both of them) we found that the limits of the denominators were not zero.


b \(\mathop {\lim }\limits_{x \to - 4} \left[ {f\left( x \right)g\left( x \right)h\left( x \right)} \right]\) Show Solution

Here is the work for this limit. At each step the property (or properties) used are listed and note that in some cases the properties may have been used more than once in the indicated step.

\[\begin{alignat*}{3}\mathop {\lim }\limits_{x \to - 4} \left[ {f\left( x \right)g\left( x \right)h\left( x \right)} \right] & = \left[ {\mathop {\lim }\limits_{x \to - 4} f\left( x \right)} \right]\left[ {\mathop {\lim }\limits_{x \to - 4} g\left( x \right)} \right]\left[ {\mathop {\lim }\limits_{x \to - 4} h\left( x \right)} \right] & & \hspace{0.25in}{\mbox{Property 3}}\\ & = \left( 1 \right)\left( {10} \right)\left( { - 7} \right) & & \hspace{0.25in}{\mbox{Plug in value of limits}}\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{ - 70}} & & & \end{alignat*}\]

Note that the properties 2 & 3 in this section were only given with two functions but they can easily be extended out to more than two functions as we did here for property 3.


c \(\displaystyle \mathop {\lim }\limits_{x \to - 4} \left[ {\frac{1}{{h\left( x \right)}} + \frac{{3 - f\left( x \right)}}{{g\left( x \right) + h\left( x \right)}}} \right]\) Show Solution

Here is the work for this limit. At each step the property (or properties) used are listed and note that in some cases the properties may have been used more than once in the indicated step.

\[\begin{alignat*}{3}\mathop {\lim }\limits_{x \to - 4} \left[ {\frac{1}{{h\left( x \right)}} + \frac{{3 - f\left( x \right)}}{{g\left( x \right) + h\left( x \right)}}} \right] & = \mathop {\lim }\limits_{x \to - 4} \frac{1}{{h\left( x \right)}} + \mathop {\lim }\limits_{x \to - 4} \frac{{3 - f\left( x \right)}}{{g\left( x \right) + h\left( x \right)}} & & \hspace{0.25in}{\mbox{Property 2}}\\ & = \frac{{\mathop {\lim }\limits_{x \to - 4} 1}}{{\mathop {\lim }\limits_{x \to - 4} h\left( x \right)}} + \frac{{\mathop {\lim }\limits_{x \to - 4} \left[ {3 - f\left( x \right)} \right]}}{{\mathop {\lim }\limits_{x \to - 4} \left[ {g\left( x \right) + h\left( x \right)} \right]}} & & \hspace{0.25in}{\mbox{Property 4}}\\ & = \frac{{\mathop {\lim }\limits_{x \to - 4} 1}}{{\mathop {\lim }\limits_{x \to - 4} h\left( x \right)}} + \frac{{\mathop {\lim }\limits_{x \to - 4} 3 - \mathop {\lim }\limits_{x \to - 4} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to - 4} g\left( x \right) + \mathop {\lim }\limits_{x \to - 4} h\left( x \right)}} & & \hspace{0.25in}{\mbox{Property 2}}\\ & = \frac{1}{{ - 7}} + \frac{{3 - 1}}{{10 - 7}} & & \hspace{0.25in}{\mbox{Plug in values of limits }}\\ & & & \hspace{0.25in}{\mbox{& Property 1}}\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{11}}{{21}}}} & & & \end{alignat*}\]

Note that were able to use Property 4 in the second step only because after we evaluated the limit of the denominators (both of them) we found that the limits of the denominators were not zero.


d \(\displaystyle \mathop {\lim }\limits_{x \to - 4} \left[ {2h\left( x \right) - \frac{1}{{h\left( x \right) + 7f\left( x \right)}}} \right]\) Show Solution

Here is the work for this limit. At each step the property (or properties) used are listed and note that in some cases the properties may have been used more than once in the indicated step.

\[\begin{alignat*}{3}\mathop {\lim }\limits_{x \to - 4} \left[ {2h\left( x \right) - \frac{1}{{h\left( x \right) + 7f\left( x \right)}}} \right] & = \mathop {\lim }\limits_{x \to - 4} 2h\left( x \right) - \mathop {\lim }\limits_{x \to - 4} \frac{1}{{h\left( x \right) + 7f\left( x \right)}} & & \hspace{0.25in}{\mbox{Property 2}}\\ & = \mathop {\lim }\limits_{x \to - 4} 2h\left( x \right) - \frac{{\mathop {\lim }\limits_{x \to - 4} 1}}{{\mathop {\lim }\limits_{x \to - 4} \left[ {h\left( x \right) + 7f\left( x \right)} \right]}} & & \hspace{0.25in}{\mbox{Property 4}} & & & \end{alignat*}\]

At this point let’s step back a minute. In the previous parts we didn’t worry about using property 4 on a rational expression. However, in this case let’s be a little more careful. We can only use property 4 if the limit of the denominator is not zero. Let’s check that limit and see what we get.

\[\begin{alignat*}{3}\hspace{0.25in}\mathop {\lim }\limits_{x \to - 4} \left[ {h\left( x \right) + 7f\left( x \right)} \right] & = \mathop {\lim }\limits_{x \to - 4} h\left( x \right) + \mathop {\lim }\limits_{x \to - 4} \left[ {7f\left( x \right)} \right] & & \hspace{0.25in}{\mbox{Property 2}}\\ & = \mathop {\lim }\limits_{x \to - 4} h\left( x \right) + 7\mathop {\lim }\limits_{x \to - 4} f\left( x \right) & & \hspace{0.25in}{\mbox{Property 1}}\\ & = - 7 + 7\left( 1 \right) & & \hspace{0.25in}{\mbox{Plug in values of limits & Property 1}}\\ & = 0 & & & \end{alignat*}\]

Okay, we can see that the limit of the denominator in the second term will be zero so we cannot actually use property 4 on that term. This means that this limit cannot be done and note that the fact that we could determine a value for the limit of the first term will not change this fact. This limit cannot be done.