Two methods? Evaluate the following limits in two different ways: Use the methods 5 of Chapter 2 and use l'Hôpital's Rule. lim 2x ? x +1 x?? 5x + x
Solution Step 1 5 In this problem we have to find the value of lim 2x 6 x +1in two different ways. x 5x + x The first method is to simplify the given problem and then apply the limit and the second method is to use l’hopital’s rule. l'Hôpital's Rule: f(x) 0 f(x) ± Suppose that we have one of the following cases, lim g(x) = 0 or lim g(x) = ± xa xa Where a can be any real number, infinity or negative infinity. f(x) f(x) In these cases we have lim = lim xa g(x) xa g(x) Step 2 5 Let us find the value of lim 2x 6 x +1using first method. x 5x + x So let us simplify the expression. 2x x +1 Consider lim x 5x + x 6 Divide the numerator and denominator by x , we get 2x x+1 2x x +1 x6 lim 5x + x = lim 5x +x x x x6 x 5 16 = lim x 1x x 5+x5 Now substituting limit x , we get, + = 1 5+ = 0 5 = 0 Thus by using first method we get, 2x x +1 lim 6 = 0 x 5x + x
