Solved Problems on Limits and Continuity

Mika Seppälä: Limits and Continuity Overview of Problems 1 2 4 3 5 6 7 8 9 10 Mika Seppälä: Limits and Continuity

Mika Seppälä: Limits and Continuity Overview of Problems 11 12 13 14 15 Mika Seppälä: Limits and Continuity

Main Methods of Limit Computations 1 The following undefined quantities cause problems: 2 In the evaluation of expressions, use the rules If the function, for which the limit needs to be computed, is defined by an algebraic expression, which takes a finite value at the limit point, then this finite value is the limit value. 3 If the function, for which the limit needs to be computed, cannot be evaluated at the limit point (i.e. the value is an undefined expression like in (1)), then find a rewriting of the function to a form which can be evaluated at the limit point. 4 Mika Seppälä: Limits and Continuity

Main Computation Methods Frequently needed rule 1 Cancel out common factors of rational functions. 2 If a square root appears in the expression, then multiply and divide by the conjugate of the square root expression. 3 Use the fact that 4 Mika Seppälä: Limits and Continuity

Continuity of Functions Functions defined by algebraic or elementary expressions involving polynomials, rational functions, trigonometric functions, exponential functions or their inverses are continuous at points where they take a finite well defined value. 1 A function f is continuous at a point x = a if 2 Used to show that equations have solutions. The following are not continuous x = 0: 3 4 Intermediate Value Theorem for Continuous Functions If f is continuous, f(a) < 0 and f(b) > 0, then there is a point c between a and b so that f(c) = 0. Mika Seppälä: Limits and Continuity

Mika Seppälä: Limits and Continuity Limits by Rewriting Problem 1 Solution Mika Seppälä: Limits and Continuity

Mika Seppälä: Limits and Continuity Limits by Rewriting Problem 2 Solution Mika Seppälä: Limits and Continuity

Mika Seppälä: Limits and Continuity Limits by Rewriting Problem 3 Solution Rewrite Mika Seppälä: Limits and Continuity

Mika Seppälä: Limits and Continuity Limits by Rewriting Problem 4 Solution Rewrite Next divide by x. Mika Seppälä: Limits and Continuity

Mika Seppälä: Limits and Continuity Limits by Rewriting Problem 5 Solution Rewrite Next divide by x. Mika Seppälä: Limits and Continuity

Mika Seppälä: Limits and Continuity Limits by Rewriting Problem 6 Solution Mika Seppälä: Limits and Continuity

Mika Seppälä: Limits and Continuity Limits by Rewriting Problem 7 Solution Rewrite: Mika Seppälä: Limits and Continuity

Mika Seppälä: Limits and Continuity Limits by Rewriting Problem 8 Solution Rewrite: Mika Seppälä: Limits and Continuity

Mika Seppälä: Limits and Continuity Limits by Rewriting Problem 9 Solution Rewrite Mika Seppälä: Limits and Continuity

Mika Seppälä: Limits and Continuity Limits by Rewriting Problem 9 Solution (cont’d) Rewrite Next divide by x. Here we used the fact that all sin(x)/x terms approach 1 as x  0. Mika Seppälä: Limits and Continuity

Mika Seppälä: Limits and Continuity One-sided Limits Problem 10 Solution Mika Seppälä: Limits and Continuity

Mika Seppälä: Limits and Continuity Problem 11 Solution Mika Seppälä: Limits and Continuity

Mika Seppälä: Limits and Continuity Problem 12 Solution Mika Seppälä: Limits and Continuity

Mika Seppälä: Limits and Continuity Problem 13 Solution Mika Seppälä: Limits and Continuity

Mika Seppälä: Limits and Continuity Problem 14 Answer Removable Not removable Mika Seppälä: Limits and Continuity

Mika Seppälä: Limits and Continuity Problem 15 Solution By the intermediate Value Theorem, a continuous function takes any value between any two of its values. I.e. it suffices to show that the function f changes its sign infinitely often. Mika Seppälä: Limits and Continuity