Question

1. Given that

\(\mathop {lim}\limits_{x \to 2} f\left( x \right) = 4\) \(\mathop {lim}\limits_{x \to 2} g\left( x \right){\bf{ = }} - {\bf{2}}\) \(\mathop {lim}\limits_{x \to 2} h\left( x \right) = 0\)

find the limits that exist. If the limit does not exist, explainwhy.

(a)\(\mathop {lim}\limits_{x \to 2} \left( {f\left( x \right) + 5g\left( x \right)} \right)\)

(b) \(\mathop {lim}\limits_{x \to 2} {\left( {g\left( x \right)} \right)^{\bf{3}}}\)

(c) \(\mathop {lim}\limits_{x \to 2} \sqrt {f\left( x \right)} \)

(d) \(\mathop {lim}\limits_{x \to 2} \frac{{{\bf{3}}f\left( x \right)}}{{g\left( x \right)}}\)

(e) \(\mathop {lim}\limits_{x \to 2} \frac{{g\left( x \right)}}{{h\left( x \right)}}\)

(f) \(\mathop {lim}\limits_{x \to 2} \frac{{g\left( x \right)h\left( x \right)}}{{f\left( x \right)}}\)

Short answer

1. \(\mathop {lim}\limits_{x \to 2} \left( {f\left( x \right) + 5g\left( x \right)} \right) = - 6\)
2. \(\mathop {lim}\limits_{x \to 2} {\left( {g\left( x \right)} \right)^3} = - 8\)
3. \(\mathop {lim}\limits_{x \to 2} \sqrt {f\left( x \right)} = 2\)
4. \(\mathop {lim}\limits_{x \to 2} \frac{{3f\left( x \right)}}{{g\left( x \right)}} = - 6\)
5. \(\mathop {lim}\limits_{x \to 2} \frac{{g\left( x \right)}}{{h\left( x \right)}} = {\rm{does}}\;{\rm{not}}\;{\rm{exist}}\)
6. \(\mathop {lim}\limits_{x \to 2} \frac{{g\left( x \right)h\left( x \right)}}{{f\left( x \right)}} = 0\)

Step-by-step solution

Apply Sum laws

According to 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 the given function as:

\(\mathop {lim}\limits_{x \to 2} \left( {f\left( x \right) + 5g\left( x \right)} \right) = \mathop {lim}\limits_{x \to 2} f\left( x \right) + \mathop {lim}\limits_{x \to 2} \left( {5g\left( x \right)} \right)\)

(a)Step 2: Apply Constant Multiple laws

According to the constant multiple law, \(\mathop {\lim }\limits_{x \to a} cf\left( x \right) = c\mathop {\lim }\limits_{x \to a} f\left( x \right)\), where \(\mathop {\lim }\limits_{x \to a} f\left( x \right)\) exists.

Apply the law and use the given values as shown below:

\(\begin{array}{c}\mathop {lim}\limits_{x \to 2} f\left( x \right) + \mathop {lim}\limits_{x \to 2} \left( {5g\left( x \right)} \right) &=& \mathop {lim}\limits_{x \to 2} f\left( x \right) + 5\mathop {lim}\limits_{x \to 2} g\left( x \right)\\ &=& 4 + 5\left( { - 2} \right)\\ &=& 4 - 10\\ &=& - 6\end{array}\)

Thus, the value of the limit is \( - 6\).

(b) Step 3: Apply Power law

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

Apply the law and use given values as:

\(\begin{array}{c}\mathop {lim}\limits_{x \to 2} {\left( {g\left( x \right)} \right)^3} &=& {\left( {\mathop {lim}\limits_{x \to 2} g\left( x \right)} \right)^3}\\ &=& {\left( { - 2} \right)^3}\\ &=& - 8\end{array}\)

Thus, the value of the limit is \( - 8\).

(c)  Step 4: Apply Power laws

According to the Root law, \(\mathop {\lim }\limits_{x \to a} \sqrt(n){{f\left( x \right)}} = \sqrt(n){{\mathop {\lim }\limits_{x \to a} f\left( x \right)}}\), where \(\mathop {\lim }\limits_{x \to a} f\left( x \right)\) exists and \(n\) is a positive integer.

Apply the law and use the given values as:

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

Thus, the value of the limit is 2.

(d) Step 5: Apply Quotient laws

According to 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 \(\mathop {\lim }\limits_{x \to a} g\left( x \right) \ne 0\).

Apply the law and use the given values as:

\(\begin{array}{c}\mathop {lim}\limits_{x \to 2} \frac{{3f\left( x \right)}}{{g\left( x \right)}} &=& \frac{{\mathop {lim}\limits_{x \to 2} 3f\left( x \right)}}{{\mathop {lim}\limits_{x \to 2} g\left( x \right)}}\\ &=& \frac{{3\mathop {lim}\limits_{x \to 2} f\left( x \right)}}{{\mathop {lim}\limits_{x \to 2} g\left( x \right)}}\\ &=& \frac{{3\left( 4 \right)}}{{ - 2}}\\ &=& - 6\end{array}\)

Thus, the value of the limit is \( - 6\).

(e)  Step 6: Apply Quotient laws

According to 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 \(\mathop {\lim }\limits_{x \to a} g\left( x \right) \ne 0\).

Since the limit of \(\mathop {lim}\limits_{x \to 2} h\left( x \right)\) is \(0\), therefore we cannot use quotient law.

Hence, the limit of the function \(\mathop {lim}\limits_{x \to 2} \frac{{g\left( x \right)}}{{h\left( x \right)}}\) does not exist as the denominator tends to \(0\).

(f)  Step 7: Apply Quotient laws

According to 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 \(\mathop {\lim }\limits_{x \to a} g\left( x \right) \ne 0\).

Apply the law and use given values as:

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

Apply Product laws

According to 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 given values as:

\(\begin{array}{c}\frac{{\mathop {lim}\limits_{x \to 2} g\left( x \right)h\left( x \right)}}{{\mathop {lim}\limits_{x \to 2} f\left( x \right)}} &=& \frac{{\mathop {lim}\limits_{x \to 2} g\left( x \right) \cdot \mathop {lim}\limits_{x \to 2} h\left( x \right)}}{{\mathop {lim}\limits_{x \to 2} f\left( x \right)}}\\ &=& \frac{{\left( { - 2} \right) \cdot \left( 0 \right)}}{4}\\ &=& 0\end{array}\)

Thus, the value of the limit is 0.