Question

The graphs of \(f\) and \(g\) are given. Use them to evaluate each

limit, if it exists. If the limit does not exist, explain why.

(a) \(\mathop {lim}\limits_{x \to 2} \left( {f\left( x \right) + g\left( x \right)} \right)\)

(b) \(\mathop {lim}\limits_{x \to 0} \left( {f\left( x \right) - g\left( x \right)} \right)\)

(c) \(\mathop {lim}\limits_{x \to - 1} \left( {f\left( x \right)g\left( x \right)} \right)\)

(d) \(\mathop {lim}\limits_{x \to 3} \frac{{f\left( x \right)}}{{g\left( x \right)}}\)

(e) \(\mathop {lim}\limits_{x \to 2} \left( {{x^2}f\left( x \right)} \right)\)

(f) \(f\left( { - {\bf{1}}} \right) + \mathop {lim}\limits_{x \to - {\bf{1}}} g\left( x \right)\)

Short answer

(a) \(\mathop {lim}\limits_{x \to 2} \left( {f\left( x \right) + g\left( x \right)} \right) = 1\)

(b) \(\mathop {lim}\limits_{x \to 0} \left( {f\left( x \right) - g\left( x \right)} \right) = {\rm{does}}\;{\rm{not}}\;{\rm{exist}}\)

(c) \(\mathop {lim}\limits_{x \to 1} \left( {f\left( x \right)g\left( x \right)} \right) = 2\)

(d) \(\mathop {lim}\limits_{x \to 3} \frac{{f\left( x \right)}}{{g\left( x \right)}} = {\rm{does}}\;{\rm{not}}\;{\rm{exist}}\)

(e) \(\mathop {lim}\limits_{x \to 2} \left( {{x^2}f\left( x \right)} \right) = - 4\)

(f) \(f\left( { - 1} \right)\mathop {lim}\limits_{x \to - 1} g\left( x \right) = {\rm{does}}\;{\rm{not}}\;{\rm{exist}}\)

Step-by-step solution

Write the limits of function

It is observed from the graph that, \(\mathop {\lim }\limits_{x \to 2} f\left( x \right) = - 1\), \(\mathop {\lim }\limits_{x \to 2} g\left( x \right) = 2\), \(\mathop {\lim }\limits_{x \to 0} g\left( x \right) = {\rm{does}}\;{\rm{not}}\,{\rm{exist}}\), \(\mathop {\lim }\limits_{x \to - 1} f\left( x \right) = 1\), \(\mathop {\lim }\limits_{x \to - 1} g\left( x \right) = 2\), \(\mathop {\lim }\limits_{x \to 3} f\left( x \right) = 1\) and \(\mathop {\lim }\limits_{x \to 3} g\left( x \right) = 0\).

(a) Step 2: Apply Sum laws

According the sum law, \(\mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) + g\left( x \right)} \right) = \mathop {\lim }\limits_{x \to a} f\left( x \right) + \mathop {\lim }\limits_{x \to a} g\left( x \right)\), where \(\mathop {\lim }\limits_{x \to a} f\left( x \right)\) and \(\mathop {\lim }\limits_{x \to a} g\left( x \right)\) exists.

Apply the law in given function and substitute the obtained values as:

\(\begin{array}{c}\mathop {lim}\limits_{x \to 2} \left( {f\left( x \right) + g\left( x \right)} \right) &=& \mathop {lim}\limits_{x \to 2} f\left( x \right) + \mathop {lim}\limits_{x \to 2} \left( {g\left( x \right)} \right)\\ &=& - 1 + 2\\ &=& 1\end{array}\)

Thus, the value of the limit is 1.

(b) Step 3: Apply Difference laws

According the Difference law, \(\mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) - g\left( x \right)} \right) = \mathop {\lim }\limits_{x \to a} f\left( x \right) - \mathop {\lim }\limits_{x \to a} g\left( x \right)\), where \(\mathop {\lim }\limits_{x \to a} f\left( x \right)\) and \(\mathop {\lim }\limits_{x \to a} g\left( x \right)\) exists.

From the graph, it is observed that \(\mathop {\lim }\limits_{x \to 0} g\left( x \right)\) does not exist. This implies that, difference law cannot be applied.

Therefore, \(\mathop {lim}\limits_{x \to 0} \left( {f\left( x \right) - g\left( x \right)} \right)\) does not exist.

(c) Step 4: Apply Product laws

According the product law, \(\mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) \cdot g\left( x \right)} \right) = \mathop {\lim }\limits_{x \to a} f\left( x \right) \cdot \mathop {\lim }\limits_{x \to a} g\left( x \right)\), where \(\mathop {\lim }\limits_{x \to a} f\left( x \right)\) and \(\mathop {\lim }\limits_{x \to a} g\left( x \right)\) exists.

Apply the law and use the values as shown below:

\(\begin{array}{c}\mathop {lim}\limits_{x \to 1} \left( {f\left( x \right)g\left( x \right)} \right) &=& \mathop {lim}\limits_{x \to 1} f\left( x \right) \cdot \mathop {lim}\limits_{x \to 1} g\left( x \right)\\ &=& 1 \cdot 2\\ &=& 2\end{array}\)

Thus, the value is 2.

(d) Step 5: Apply Quotient laws

According the Quotient law, \(\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{f\left( x \right)}} = \frac{{\mathop {\lim }\limits_{x \to a} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to a} g\left( x \right)}}\), where \(\mathop {\lim }\limits_{x \to a} f\left( x \right)\) and \(\mathop {\lim }\limits_{x \to a} g\left( x \right)\) exists and \(\mathop {\lim }\limits_{x \to a} g\left( x \right) \ne 0\).

We cannot apply quotient rule because the value of \(\mathop {\lim }\limits_{x \to 3} g\left( x \right) = 0\). As, \(\mathop {lim}\limits_{x \to {3^ - }} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \infty \) and \(\mathop {lim}\limits_{x \to {3^ + }} \frac{{f\left( x \right)}}{{g\left( x \right)}} = - \infty \).

This implies that \(\mathop {lim}\limits_{x \to 3} \frac{{f\left( x \right)}}{{g\left( x \right)}}\) does not exist.

(e) Step 6: Apply Product laws

According the product law, \(\mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) \cdot g\left( x \right)} \right) = \mathop {\lim }\limits_{x \to a} f\left( x \right) \cdot \mathop {\lim }\limits_{x \to a} g\left( x \right)\), where \(\mathop {\lim }\limits_{x \to a} f\left( x \right)\) and \(\mathop {\lim }\limits_{x \to a} g\left( x \right)\) exists.

Apply the law as:

\(\mathop {lim}\limits_{x \to 2} \left( {{x^2}f\left( x \right)} \right) = \mathop {lim}\limits_{x \to 2} \left( {{x^2}} \right) \cdot \mathop {lim}\limits_{x \to 2} f\left( x \right)\)

Apply limit laws

According to the law, \(\mathop {\lim }\limits_{x \to a} c = c\), and \(\mathop {\lim }\limits_{x \to a} {x^n} = {a^n}\)where \(n\) is positive integer.

Apply the law and use the values as;

\(\begin{array}{c}\mathop {lim}\limits_{x \to 2} \left( {{x^2}} \right) \cdot \mathop {lim}\limits_{x \to 2} f\left( x \right) &=& {\left( 2 \right)^2} \cdot \left( { - 1} \right)\\ &=& 4 \cdot \left( { - 1} \right)\\ &=& - 4\end{array}\)

Thus, the value of the limit is \( - 4\).

(f) Step 8: Apply limit laws

From the graph the value of \(f\left( { - 1} \right)\) is not defined.

Therefore, the value of \(f\left( { - 1} \right) + \mathop {lim}\limits_{x \to - 1} g\left( x \right)\) does not exists.