Section 2.5 : Computing Limits
For problems 1 – 20 evaluate the limit, if it exists.
- \(\mathop {\lim }\limits_{x \to \, - 9} \left( {1 - 4{x^3}} \right)\)
- \(\mathop {\lim }\limits_{y \to 1} \left( {6{y^4} - 7{y^3} + 12y + 25} \right)\)
- \(\displaystyle \mathop {\lim }\limits_{t \to 0} \frac{{{t^2} + 6}}{{{t^2} - 3}}\)
- \(\displaystyle \mathop {\lim }\limits_{z \to 4} \frac{{6z}}{{2 + 3{z^2}}}\)
- \(\displaystyle \mathop {\lim }\limits_{w \to \, - 2} \frac{{w + 2}}{{{w^2} - 6w - 16}}\)
- \(\displaystyle \mathop {\lim }\limits_{t \to - 5} \frac{{{t^2} + 6t + 5}}{{{t^2} + 2t - 15}}\)
- \(\displaystyle \mathop {\lim }\limits_{x \to 3} \frac{{5{x^2} - 16x + 3}}{{9 - {x^2}}}\)
- \(\displaystyle \mathop {\lim }\limits_{z \to 1} \frac{{10 - 9z - {z^2}}}{{3{z^2} + 4z - 7}}\)
- \(\displaystyle \mathop {\lim }\limits_{x \to \, - 2} \frac{{{x^3} + 8}}{{{x^2} + 8x + 12}}\)
- \(\displaystyle \mathop {\lim }\limits_{t \to 8} \frac{{t\left( {t - 5} \right) - 24}}{{{t^2} - 8t}}\)
- \(\displaystyle \mathop {\lim }\limits_{w \to \, - 4} \frac{{{w^2} - 16}}{{\left( {w - 2} \right)\left( {w + 3} \right) - 6}}\)
- \(\displaystyle \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {2 + h} \right)}^3} - 8}}{h}\)
- \(\displaystyle \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {1 + h} \right)}^4} - 1}}{h}\)
- \(\displaystyle \mathop {\lim }\limits_{t \to 25} \frac{{5 - \sqrt t }}{{t - 25}}\)
- \(\displaystyle \mathop {\lim }\limits_{x \to \,2} \frac{{x - 2}}{{\sqrt 2 - \sqrt x }}\)
- \(\displaystyle \mathop {\lim }\limits_{z \to 6} \frac{{z - 6}}{{\sqrt {3z - 2} - 4}}\)
- \(\displaystyle \mathop {\lim }\limits_{z \to \, - 2} \frac{{3 - \sqrt {1 - 4z} }}{{2z + 4}}\)
- \(\displaystyle \mathop {\lim }\limits_{t \to 3} \frac{{3 - t}}{{\sqrt {t + 1} - \sqrt {5t - 11} }}\)
- \(\displaystyle \mathop {\lim }\limits_{x \to 7} \frac{{\,\frac{1}{7} - \frac{1}{x}\,}}{{x - 7}}\)
- \(\displaystyle \mathop {\lim }\limits_{y \to \, - 1} \frac{{\frac{1}{{4 + 3y}} + \frac{1}{y}}}{{y + 1}}\)
- Given the function
\[f\left( x \right) = \left\{ {\begin{align*}{15}&{\hspace{0.25in}x < - 4}\\{6 - 2x}&{\hspace{0.25in}x \ge - 4}\end{align*}} \right.\]
Evaluate the following limits, if they exist.
- \(\mathop {\lim }\limits_{x \to \, - 7} f\left( x \right)\)
- \(\mathop {\lim }\limits_{x \to - 4} f\left( x \right)\)
- Given the function
\[g\left( t \right) = \left\{ {\begin{align*}{{t^2} - {t^3}}&{\hspace{0.25in}t < 2}\\{5t - 14}&{\hspace{0.25in}t \ge 2}\end{align*}} \right.\]
Evaluate the following limits, if they exist.
- \(\mathop {\lim }\limits_{t \to - 3} g\left( t \right)\)
- \(\mathop {\lim }\limits_{t \to 2} g\left( t \right)\)
- Given the function
\[h\left( w \right) = \left\{ {\begin{align*}{2{w^2}}&{\hspace{0.25in}w \le 6}\\{w - 8}&{\hspace{0.25in}w > 6}\end{align*}} \right.\]
Evaluate the following limits, if they exist.
- \(\mathop {\lim }\limits_{w \to 6} h\left( w \right)\)
- \(\mathop {\lim }\limits_{w \to 2} h\left( w \right)\)
- Given the function
\[g\left( x \right) = \left\{ {\begin{array}{rc}{5x + 24}&{x < - 3}\\{{x^2}}&{ - 3 \le x < 4}\\{1 - 2x}&{x \ge 4}\end{array}} \right.\]
Evaluate the following limits, if they exist.
- \(\mathop {\lim }\limits_{x \to \, - 3} g\left( x \right)\)
- \(\mathop {\lim }\limits_{x \to \,0} g\left( x \right)\)
- \(\mathop {\lim }\limits_{x \to \,4} g\left( x \right)\)
- \(\mathop {\lim }\limits_{x \to \,12} g\left( x \right)\)
For problems 25 – 30 evaluate the limit, if it exists.
- \(\mathop {\lim }\limits_{z \to \, - 10} \left( {\left| {t + 10} \right| + 3} \right)\)
- \(\mathop {\lim }\limits_{x \to 4} \left( {9 + \left| {8 - 2x} \right|} \right)\)
- \(\displaystyle \mathop {\lim }\limits_{h \to 0} \frac{{\left| h \right|}}{h}\)
- \(\displaystyle \mathop {\lim }\limits_{t \to 2} \frac{{2 - t}}{{\left| {t - 2} \right|}}\)
- \(\displaystyle \mathop {\lim }\limits_{w \to \, - 5} \frac{{\left| {2w + 10} \right|}}{{w + 5}}\)
- \(\displaystyle \mathop {\lim }\limits_{x \to 4} \frac{{\left| {x - 4} \right|}}{{{x^2} - 16}}\)
- Given that \(3 + 2x \le f\left( x \right) \le x - 1\) for all x determine the value of \(\mathop {\lim }\limits_{x \to - 4} f\left( x \right)\).
- Given that \(\sqrt {x + 7} \le f\left( x \right) \le \frac{{x - 1}}{2}\) for all x determine the value of \(\mathop {\lim }\limits_{x \to 9} f\left( x \right)\).
- Use the Squeeze Theorem to determine the value of \(\displaystyle \mathop {\lim }\limits_{x \to 0} {x^4}\cos \left( {\frac{3}{x}} \right)\).
- Use the Squeeze Theorem to determine the value of \(\displaystyle \mathop {\lim }\limits_{x \to 0} x\cos \left( {\frac{1}{x}} \right)\).
- Use the Squeeze Theorem to determine the value of \(\displaystyle \mathop {\lim }\limits_{x \to 1} {\left( {x - 1} \right)^2}\cos \left( {\frac{1}{{x - 1}}} \right)\).