Allan Nambafu - DIFFERENTIATION BASICS: Calculus Hand Book by Allan Mabele Nambafu

Differentiation 1 LIMITS AND CONTINUITY Let f(x) be a function defined on an interval that contains x = ax = a, except possibly at x = ax = a. Then we say that, (lim ( x) = L a Examples 1 Find the limits of the following Solution We need to factor both numerator and denominator as shown below. The simplify to obtain, Example 2. Calculate the limit We need to look at the limit from the left of 2 and the limit from the right of 2. As x approaches 2 from the left x - 2 < 0 hence |x - 2| = - (x - 2) Substitute to obtain the limit from the left of 2 as follows Differentiation 2 = - 8 As x approaches 2 from the right x - 2 > 0 hence |x - 2| = x - 2 Substitute to obtain the limit from the right of 2 as follows = 8 The limit from the right of 2 and the limit from the left of 2 are not equal therefore the given limit DOES NOT EXIST. infinity Let us rewrite the limit so that it is of the infinity/infinity indeterminate form. Differentiation 3 We now use L'hopital's Rule and find the limit. Example 5: Find the limit Solution to Example 5: As x gets larger x + 1 gets larger and e^(1/(x+1)-1) approaches 0 hence an indeterminate form infinity.0 Let us rewrite the limit so that it is of the 0/0 indeterminate form. Example 5: Find the limit Solution to Example 5: As x gets larger x + 1 gets larger and e^(1/(x+1)-1) approaches 0 hence an indeterminate form infinity.0 Let us rewrite the limit so that it is of the 0/0 indeterminate form.

Apply the l'hopital's theorem to find the limit. Differentiation 4 = - 1 Example 5: Find the limit Solution As x approaches 9, both numerator and denominator approach 0. Multiply both numerator and denominator by the conjugate of the numerator. Expand and simplify. and now find the limit. -1 <= cos x <= 1 Divide all terms of the above inequality by x, for x positive. -1 / x <= cos x / x <= 1 / x Now as x takes larger values without bound (+infinity) both -1 / x and 1 / x approaches 0. -1 / x <= cos x / x <= 1 / x Now as x takes larger values without bound (+infinity) both -1 / x and 1 / x approaches 0.

Hence by the squeezing theorem the above limit is given by Example 7: Find the limit Solution As t approaches 0, both the numerator and denominator approach 0 and we have the 0 / 0 indeterminate form. Hence the l'hopital theorem is used to calculate the above limit as follows Differentiation 6 Example 8: Find the limit Solution We first factor out 16 x 2 under the square root of the denominator and take out of the square root and rewrite the limit as Since x approaches larger positive values (infinity) | x | = x. Simplify and find the limt. = 3 / 4 Example 9: Find the limit Solution Factor x 2 in the denominator and simplify. Differentiation 7 As x takes large values (infinity), the terms 2/x and 1/x 2 approaches 0 hence the limit is = 3 / 4 Example 10: Find the limit Solution Factor x 2 in the numerator and denominator and simplify. As x takes large values (infinity), the terms 1/x and 1/x 2 and 3/x 2 approaches 0 hence the limit is = 0 / 2 = 0