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Section 2.5 : Computing Limits
For problems 1 – 9 evaluate the limit, if it exists.
- \(\mathop {\lim }\limits_{x \to 2} \left( {8 - 3x + 12{x^2}} \right)\) Solution
- \(\displaystyle \mathop {\lim }\limits_{t \to \, - 3} \frac{{6 + 4t}}{{{t^2} + 1}}\) Solution
- \(\displaystyle \mathop {\lim }\limits_{x \to \, - 5} \frac{{{x^2} - 25}}{{{x^2} + 2x - 15}}\) Solution
- \(\displaystyle \mathop {\lim }\limits_{z \to 8} \frac{{2{z^2} - 17z + 8}}{{8 - z}}\) Solution
- \(\displaystyle \mathop {\lim }\limits_{y \to 7} \frac{{{y^2} - 4y - 21}}{{3{y^2} - 17y - 28}}\) Solution
- \(\displaystyle \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {6 + h} \right)}^2} - 36}}{h}\) Solution
- \(\displaystyle \mathop {\lim }\limits_{z \to 4} \frac{{\sqrt z - 2}}{{z - 4}}\) Solution
- \(\displaystyle \mathop {\lim }\limits_{x \to \, - 3} \frac{{\sqrt {2x + 22} - 4}}{{x + 3}}\) Solution
- \(\displaystyle \mathop {\lim }\limits_{x \to 0} \frac{x}{{3 - \sqrt {x + 9} }}\) Solution
- Given the function
\[f\left( x \right) = \left\{ {\begin{array}{rc}{7 - 4x}&{x < 1}\\{{x^2} + 2}&{x \ge 1}\end{array}} \right.\]
Evaluate the following limits, if they exist.
- \(\mathop {\lim }\limits_{x \to \, - 6} f\left( x \right)\)
- \(\mathop {\lim }\limits_{x \to 1} f\left( x \right)\)
- Given
\[h\left( z \right) = \left\{ {\begin{array}{rc}{6z}&{z \le - 4}\\{1 - 9z}&{z > - 4}\end{array}} \right.\]
Evaluate the following limits, if they exist.
- \(\mathop {\lim }\limits_{z \to 7} h\left( z \right)\)
- \(\mathop {\lim }\limits_{z \to - 4} h\left( z \right)\)
For problems 12 & 13 evaluate the limit, if it exists.
- \(\mathop {\lim }\limits_{x \to 5} \left( {10 + \left| {x - 5} \right|} \right)\) Solution
- \(\displaystyle \mathop {\lim }\limits_{t \to \, - 1} \frac{{t + 1}}{{\left| {t + 1} \right|}}\) Solution
- Given that \(x^{3}-6x^{2}+12x-3 \le f\left( x \right) \le x^{2}-4x+9\) for \(x \le 3\) determine the value of \(\mathop {\lim }\limits_{x \to 2} f\left( x \right)\). Solution
- Use the Squeeze Theorem to determine the value of \(\displaystyle \mathop {\lim }\limits_{x \to 0} {x^4}\sin \left( {\frac{\pi }{x}} \right)\). Solution