Let f(x) and g(x) be functions of x. k, c and a be contants.
1. If f(x) = c, then \(\mathop {\lim }\limits_{x \to a} f(x) = c\)
Let \(\mathop {\lim }\limits_{x \to a} f(x) = A\) and \(\mathop {\lim }\limits_{x \to a} g(x) = B\) then;
2. \(\mathop {\lim }\limits_{x \to a} kf(x) = k\mathop {\lim }\limits_{x \to a} f(x) =kA \)
3. \(\mathop {\lim }\limits_{x \to a} \left[ {f(x) \pm g(x)} \right] = \mathop {\lim }\limits_{x \to a} \left[ {f(x)} \right] \pm \mathop {\lim }\limits_{x \to a} \left[ {g(x)} \right] = A \pm B\)
4. \(\mathop {\lim }\limits_{x \to a} \left[ {f(x)g(x)} \right] = \left( {\mathop {\lim }\limits_{x \to a} f(x)} \right)\left( {\mathop {\lim }\limits_{x \to a} g(x)} \right) = AB\)
5. \(\mathop {\lim }\limits_{x \to a} \frac{{f(x)}}{{g(x)}} = \frac{{\mathop {\lim }\limits_{x \to a} f(x)}}{{\mathop {\lim }\limits_{x \to a} g(x)}} = \frac{A}{B}\) , B is not equal 0.
6. L' Hospital's Rule
7. \(\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1\)
How to calculate limits
Sample calculations
Calculate the limits of the following expressions.
1. \(\mathop {\lim }\limits_{x \to 1} \frac{{({x^2} - 1)}}{{(x - 1)}}\)\(\mathop {\lim }\limits_{x \to 1} \frac{{({x^2} - 1)}}{{(x - 1)}} = \mathop {\lim }\limits_{x \to 1} \frac{{(x + 1)(x - 1)}}{{(x - 1)}}\)
\(\mathop {\lim }\limits_{x \to 1} \frac{{({x^2} - 1)}}{{(x - 1)}} = \mathop {\lim }\limits_{x \to 1} (x + 1)\)
\(\mathop {\lim }\limits_{x \to 1} \frac{{({x^2} - 1)}}{{(x - 1)}} = \left( {1 + 1} \right) = 2\)
\(\mathop {\lim }\limits_{x \to 0} \frac{{\sin \pi x}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{\sin \pi x}}{x}\left( {\frac{\pi }{\pi }} \right)\)
\(\mathop {\lim }\limits_{x \to 0} \frac{{\sin \pi x}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{\pi \sin \pi x}}{{\pi x}}\)
\(\mathop {\lim }\limits_{x \to 0} \frac{{\sin \pi x}}{x} = \left( \pi \right)\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin \pi x}}{{\pi x}}} \right)\)
\(\mathop {\lim }\limits_{x \to 0} \frac{{\sin \pi x}}{x} = \pi \left( 1 \right)\)
\(\mathop {\lim }\limits_{x \to 0} \frac{{\sin \pi x}}{x} = \pi \)
\(\mathop {\lim }\limits_{\theta \to 0} \sqrt {1 - {{\sin }^2}\theta } = \sqrt {1 - \sin 0} \)
\(\mathop {\lim }\limits_{\theta \to 0} \sqrt {1 - {{\sin }^2}\theta } = \sqrt 1 \)
\(\mathop {\lim }\limits_{\theta \to 0} \sqrt {1 - {{\sin }^2}\theta } = \pm 1\)
\(\mathop {\lim }\limits_{x \to \infty } \frac{{x + 16}}{{x - 16}} = \mathop {\lim }\limits_{x \to \infty } \left( {\frac{{x + 16}}{{x - 16}}} \right)\left( {\frac{{\frac{1}{x}}}{{\frac{1}{x}}}} \right)\)
\(\mathop {\lim }\limits_{x \to \infty } \frac{{x + 16}}{{x - 16}} = \mathop {\lim }\limits_{x \to \infty } \left( {\frac{{1 + \frac{{16}}{x}}}{{1 - \frac{{16}}{x}}}} \right)\)
\(\mathop {\lim }\limits_{x \to \infty } \frac{{x + 16}}{{x - 16}} = \left( {\frac{{1 + 0}}{{1 - 0}}} \right) = 1\)
\(\mathop {\lim }\limits_{\theta \to 0} \frac{{1 - \cos \theta }}{{{\theta ^2}}} = \mathop {\lim }\limits_{\theta \to 0} \left( {\frac{{\sin \theta }}{{2\theta }}} \right)\) by L' Hospital's Rule
\(\mathop {\lim }\limits_{\theta \to 0} \frac{{1 - \cos \theta }}{{{\theta ^2}}} = \frac{1}{2}\mathop {\lim }\limits_{\theta \to 0} \left( {\frac{{\sin \theta }}{\theta }} \right)\)
\(\mathop {\lim }\limits_{\theta \to 0} \frac{{1 - \cos \theta }}{{{\theta ^2}}} = \frac{1}{2}\left( 1 \right)\)
\(\mathop {\lim }\limits_{\theta \to 0} \frac{{1 - \cos \theta }}{{{\theta ^2}}} = \frac{1}{2}\)
Solving Derivatives Using Limit Definition
\(\mathop {\lim }\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = \frac{{dy}}{{dx}} = f'(x)\)
Examples: Find the first derivative of the given functions
1. \(f(x) = y = {x^2} + 2x - 7\)
Solution\(f(x) = y = {x^2} + 2x - 7\)
\(y + \Delta y = {\left( {x + \Delta x} \right)^2} + 2\left( {x + \Delta x} \right) - 7\)
\(y + \Delta y = \left( {{x^2} + 2\Delta xx + \Delta {x^2}} \right) + 2\left( {x + \Delta x} \right) - 7\)
\(\Delta y = {x^2} + 2x - 7 + 2\Delta xx + \Delta x + \Delta {x^2} - y\)
\(\Delta y = {x^2} + 2x - 7 + 2\Delta xx + \Delta x + \Delta {x^2} - {x^2} - 2x + 7\)
\(\frac{{\Delta y}}{{\Delta x}} = \frac{{2\Delta xx + \Delta x + \Delta {x^2}}}{{\Delta x}}\)
\(\frac{{\Delta y}}{{\Delta x}} = 2x + 1 + \Delta x\)
\(\mathop {\lim }\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = \frac{{dy}}{{dx}} = \mathop {\lim }\limits_{\Delta x \to 0} \left( {2x + 1 + \Delta x} \right)\)
\(\mathop {\lim }\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = \frac{{dy}}{{dx}} = 2x + 1\)
2. \( y = {4x^2} - x - 1\)
Solution\(y = 4{x^2} - x - 1\)
\(y + \Delta y = 4{\left( {x + \Delta x} \right)^2} - \left( {x + \Delta x} \right) - 1\)
\(\Delta y = 4{\left( {x + \Delta x} \right)^2} - \left( {x + \Delta x} \right) - 1 - y\)
\(\Delta y = 4\left( {{x^2} + 2x\Delta x + \Delta {x^2}} \right) - \left( {x + \Delta x} \right) - 1 - y\)
\(\Delta y = \left( {4{x^2} + 8x\Delta x + 4\Delta {x^2}} \right) - \left( {x + \Delta x} \right) - 1 - \left( {4{x^2} - x - 1} \right)\)
\(\Delta y = \left( {8x\Delta x + 4\Delta {x^2}} \right)\)
\(\frac{{\Delta y}}{{\Delta x}} = \frac{{\left( {8x\Delta x + 4\Delta {x^2}} \right)}}{{\Delta x}}\)
\(\mathop {\lim }\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\left( {8x\Delta x + 4\Delta {x^2}} \right)}}{{\Delta x}} = \frac{{dy}}{{dx}}\)
\(\frac{{dy}}{{dx}} = \mathop {\lim }\limits_{\Delta x \to 0} \left( {8x + 4\Delta x} \right)\)
\(\frac{{dy}}{{dx}} = 8x - 1\)