Section 2.10 : The Definition of the Limit
Use the definition of the limit to prove the following limits.
- \(\mathop {\lim }\limits_{x \to - 4} \left( {2x} \right) = - 8\)
- \(\mathop {\lim }\limits_{x \to 1} \left( { - 7x} \right) = - 7\)
- \(\mathop {\lim }\limits_{x \to 3} \left( {2x + 8} \right) = 14\)
- \(\mathop {\lim }\limits_{x \to 2} \left( {5 - x} \right) = 3\)
- \(\mathop {\lim }\limits_{x \to \,\, - 2} {x^2} = 4\)
- \(\mathop {\lim }\limits_{x \to \,\,4} {x^2} = 16\)
- \(\mathop {\lim }\limits_{x \to 1} \left( {{x^2} + x + 6} \right) = 8\)
- \(\mathop {\lim }\limits_{x \to - 2} \left( {{x^2} + 3x - 1} \right) = - 3\)
- \(\mathop {\lim }\limits_{x \to \,1} {x^4} = 1\)
- \(\displaystyle \mathop {\lim }\limits_{x \to - 6} \frac{1}{{{{\left( {x + 6} \right)}^2}}} = \infty \)
- \(\displaystyle \mathop {\lim }\limits_{x \to 0} \frac{{ - 3}}{{{x^2}}} = - \infty \)
- \(\displaystyle \mathop {\lim }\limits_{x \to {0^ + }} \frac{1}{x} = \infty \)
- \(\displaystyle \mathop {\lim }\limits_{x \to {1^ - }} \frac{1}{{x - 1}} = - \infty \)
- \(\displaystyle \mathop {\lim }\limits_{x \to - \infty } \frac{1}{{{x^2}}} = 0\)
- \(\displaystyle \mathop {\lim }\limits_{x \to \infty } \frac{1}{{{x^3}}} = 0\)