[amp_mcq option1=”log x” option2=”0″ option3=”x log x” option4=”\\[\\infty \\]” correct=”option1″]<\/p>\n
The correct answer is A. log x.<\/p>\n
Let’s take a look at each option in turn.<\/p>\n
Option A: log x. This is the correct answer. As $\\alpha$ approaches 0, $x^\\alpha$ approaches 1, so the expression $\\frac{{x^\\alpha } – 1}}{\\alpha }$ approaches $\\log x$.<\/p>\n
Option B: 0. This is not the correct answer. As $\\alpha$ approaches 0, $x^\\alpha$ approaches 1, so the expression $\\frac{{x^\\alpha } – 1}}{\\alpha }$ approaches $\\log x$, which is not equal to 0.<\/p>\n
Option C: x log x. This is not the correct answer. As $\\alpha$ approaches 0, $x^\\alpha$ approaches 1, so the expression $\\frac{{x^\\alpha } – 1}}{\\alpha }$ approaches $\\log x$, which is not equal to $x \\log x$.<\/p>\n
Option D: $\\infty$. This is not the correct answer. As $\\alpha$ approaches 0, $x^\\alpha$ approaches 1, so the expression $\\frac{{x^\\alpha } – 1}}{\\alpha }$ approaches $\\log x$, which is not equal to $\\infty$.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”log x” option2=”0″ option3=”x log x” option4=”\\[\\infty \\]” correct=”option1″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[690],"tags":[],"class_list":["post-20209","post","type-post","status-publish","format-standard","hentry","category-calculus","no-featured-image-padding"],"yoast_head":"\n