The ability to evaluate a limit often depends on what’s in your “bag of tricks.” Here’s a first-year limit that uses a few:<\/p>\n
Example:<\/strong><\/p>\n Find limitx→0+<\/sub>x^x<\/p>\n Solution:<\/p>\n We can imagine<\/p>\n y=limitx→0+<\/sub>x^x<\/p>\n then use the log trick<\/p>\n lny=ln(limitx→0+<\/sub>x^x)<\/p>\n Provided everything stays defined (in this case, x>0),<\/p>\n lny=limitx→0+<\/sub>lnx^x (important!)<\/p>\n Now, we can simplify the inside right using the exponent-to-multiple-rule :<\/p>\n lny=limitx→0+<\/sub>xlnx<\/p>\n We can rewrite as<\/p>\n lny=limitx→0+<\/sub>lnx\/(1\/x)<\/p>\n which is the indeterminate form -∞\/∞ and therefore eligible for l’H\u00f4pital’s rule.<\/p>\n We proceed by taking separate derivatives top and bottom:<\/p>\n lny=limitx→0+<\/sub>(1\/x)\/(-1\/x^2)=limitx→0+<\/sub>-x=0<\/p>\n Recall from the beginning that y=limitx→0+<\/sub>x^x. Since e^y is the inverse of lny,<\/p>\n lny=0 leads to<\/p>\n y=e^0=1<\/p>\n Therefore<\/p>\n limitx→0+<\/sub> x^x = 1<\/p>\n Source:<\/p>\n Larson, Roland E. and Robert P. Hostetler. Calculus, Part One<\/u>, 3rd Edition. Toronto: \u00a0\u00a0D.C. Heath and Company, 1989.<\/p>\n