? 0 1 ,0 ,? forms ?Evaluate the following limits. Check your results by graphing. limx 1/x x??

Solution Step 1 1/x In this problem we have to evaluate limx By direct substitution of limit, we get lim x 1/x= 1/ = 0 x 1/x 0 Thus lx has the indeterminate form . To evaluate this limit we will be using the following two steps: g(x) 0, Assume limf(x) has the indeterminate form 0 1 , . 0 xa 0 1. Evaluate L = lixax)ln f(x).This limit can often be put in the form 0or both of which can be handled by l’hopital’s rule. g(x) 2. Then limf(x) = eL xa l'Hôpital's Rule: f(x) 0 f(x) ± Suppose that we have one of the following cases, lim xa g(x)= 0r lixa g(x) ± Where a can be any real number, infinity or negative infinity. f(x) f(x) In these cases we have lim g(x)= lim gx) xa xa Step 2 Here our f(x) = x,g(x) = and a = then x L = limg(x) ln f(x) x 1 ln x = lim( xln (x) = lim x = which is an indeterminate form. x x So we can apply l’hopital’s rule. ln x dxln x) lim x = lim d(x) x x dx d 1 d dx (ln x) = xnd dx(x) = 1 d 1 Thus lim ln = lim ddln x= lim = lim = 1 1 = 0 x x x dxx) x 1 x x Thus L = 0