### Math Notes

Subjects

#### Differential Calculus Solutions

##### Topics || Problems

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

$$\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{{f'\left( x \right)}}{{g'\left( x \right)}}$$

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. $$\mathop {\lim }\limits_{x \to 1} \frac{{({x^2} - 1)}}{{(x - 1)}}$$

$$\mathop {\lim }\limits_{x \to 1} \frac{{({x^2} - 1)}}{{(x - 1)}} = \mathop {\lim }\limits_{x \to 1} \frac{{(x + 1)(x - 1)}}{{(x - 1)}}$$

$$\mathop {\lim }\limits_{x \to 1} \frac{{({x^2} - 1)}}{{(x - 1)}} = \mathop {\lim }\limits_{x \to 1} (x + 1)$$

$$\mathop {\lim }\limits_{x \to 1} \frac{{({x^2} - 1)}}{{(x - 1)}} = \left( {1 + 1} \right) = 2$$

2. $$\mathop {\lim }\limits_{x \to 0} \frac{{\sin \pi x}}{x}$$

$$\mathop {\lim }\limits_{x \to 0} \frac{{\sin \pi x}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{\sin \pi x}}{x}\left( {\frac{\pi }{\pi }} \right)$$

$$\mathop {\lim }\limits_{x \to 0} \frac{{\sin \pi x}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{\pi \sin \pi x}}{{\pi x}}$$

$$\mathop {\lim }\limits_{x \to 0} \frac{{\sin \pi x}}{x} = \left( \pi \right)\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin \pi x}}{{\pi x}}} \right)$$

$$\mathop {\lim }\limits_{x \to 0} \frac{{\sin \pi x}}{x} = \pi \left( 1 \right)$$

$$\mathop {\lim }\limits_{x \to 0} \frac{{\sin \pi x}}{x} = \pi$$

3. $$\mathop {\lim }\limits_{\theta \to 0} \sqrt {1 - {{\sin }^2}\theta }$$

$$\mathop {\lim }\limits_{\theta \to 0} \sqrt {1 - {{\sin }^2}\theta } = \sqrt {1 - \sin 0}$$

$$\mathop {\lim }\limits_{\theta \to 0} \sqrt {1 - {{\sin }^2}\theta } = \sqrt 1$$

$$\mathop {\lim }\limits_{\theta \to 0} \sqrt {1 - {{\sin }^2}\theta } = \pm 1$$

4. $$\mathop {\lim }\limits_{x \to \infty } \frac{{x + 16}}{{x - 16}}$$

$$\mathop {\lim }\limits_{x \to \infty } \frac{{x + 16}}{{x - 16}} = \mathop {\lim }\limits_{x \to \infty } \left( {\frac{{x + 16}}{{x - 16}}} \right)\left( {\frac{{\frac{1}{x}}}{{\frac{1}{x}}}} \right)$$

$$\mathop {\lim }\limits_{x \to \infty } \frac{{x + 16}}{{x - 16}} = \mathop {\lim }\limits_{x \to \infty } \left( {\frac{{1 + \frac{{16}}{x}}}{{1 - \frac{{16}}{x}}}} \right)$$

$$\mathop {\lim }\limits_{x \to \infty } \frac{{x + 16}}{{x - 16}} = \left( {\frac{{1 + 0}}{{1 - 0}}} \right) = 1$$

5. $$\mathop {\lim }\limits_{\theta \to 0} \frac{{1 - \cos \theta }}{{{\theta ^2}}}$$

$$\mathop {\lim }\limits_{\theta \to 0} \frac{{1 - \cos \theta }}{{{\theta ^2}}} = \mathop {\lim }\limits_{\theta \to 0} \left( {\frac{{\sin \theta }}{{2\theta }}} \right)$$ 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. $$f(x) = y = {x^2} + 2x - 7$$

$$f(x) = y = {x^2} + 2x - 7$$

$$y + \Delta y = {\left( {x + \Delta x} \right)^2} + 2\left( {x + \Delta x} \right) - 7$$

$$y + \Delta y = \left( {{x^2} + 2\Delta xx + \Delta {x^2}} \right) + 2\left( {x + \Delta x} \right) - 7$$

$$\Delta y = {x^2} + 2x - 7 + 2\Delta xx + \Delta x + \Delta {x^2} - y$$

$$\Delta y = {x^2} + 2x - 7 + 2\Delta xx + \Delta x + \Delta {x^2} - {x^2} - 2x + 7$$

$$\frac{{\Delta y}}{{\Delta x}} = \frac{{2\Delta xx + \Delta x + \Delta {x^2}}}{{\Delta x}}$$

$$\frac{{\Delta y}}{{\Delta x}} = 2x + 1 + \Delta x$$

$$\mathop {\lim }\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = \frac{{dy}}{{dx}} = \mathop {\lim }\limits_{\Delta x \to 0} \left( {2x + 1 + \Delta x} \right)$$

$$\mathop {\lim }\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = \frac{{dy}}{{dx}} = 2x + 1$$

2. $$y = {4x^2} - x - 1$$

$$y = 4{x^2} - x - 1$$

$$y + \Delta y = 4{\left( {x + \Delta x} \right)^2} - \left( {x + \Delta x} \right) - 1$$

$$\Delta y = 4{\left( {x + \Delta x} \right)^2} - \left( {x + \Delta x} \right) - 1 - y$$

$$\Delta y = 4\left( {{x^2} + 2x\Delta x + \Delta {x^2}} \right) - \left( {x + \Delta x} \right) - 1 - y$$

$$\Delta y = \left( {4{x^2} + 8x\Delta x + 4\Delta {x^2}} \right) - \left( {x + \Delta x} \right) - 1 - \left( {4{x^2} - x - 1} \right)$$

$$\Delta y = \left( {8x\Delta x + 4\Delta {x^2}} \right)$$

$$\frac{{\Delta y}}{{\Delta x}} = \frac{{\left( {8x\Delta x + 4\Delta {x^2}} \right)}}{{\Delta x}}$$

$$\mathop {\lim }\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\left( {8x\Delta x + 4\Delta {x^2}} \right)}}{{\Delta x}} = \frac{{dy}}{{dx}}$$

$$\frac{{dy}}{{dx}} = \mathop {\lim }\limits_{\Delta x \to 0} \left( {8x + 4\Delta x} \right)$$

$$\frac{{dy}}{{dx}} = 8x - 1$$