Search for: limx→0 2|x|e|x|−|x|−|x|loge2−1xtanx is equal to limx→0 2|x|e|x|−|x|−|x|loge2−1xtanx is equal to A12(ln2)2+12(ln2)+1B(ln2)2+12(ln2)+1C(ln2)2+(ln2)+12D12(ln2)2+(ln2)+12 Fill Out the Form for Expert Academic Guidance!l Grade ---Class 1Class 2Class 3Class 4Class 5Class 6Class 7Class 8Class 9Class 10Class 11Class 12 Target Exam JEENEETCBSE +91 Preferred time slot for the call ---9 am10 am11 am12 pm1 pm2 pm3 pm4 pm5 pm6 pm7 pm8pm9 pm10pm Please indicate your interest Live ClassesBooksTest SeriesSelf Learning Language ---EnglishHindiMarathiTamilTeluguMalayalam Are you a Sri Chaitanya student? NoYes Verify OTP Code (required) I agree to the terms and conditions and privacy policy. Solution: limx→0 2|x|e|x|−|x|−|x|loge2−1xtanx=limx→0 (2e)|x|−|x|loge(2e)−1x2tanxx=limx→0 1+|x|loge(2e)+|x|22!loge(2e)2+…−|x|loge(2e)−1x2tanxx=limx→0 |x|22!loge(2e)2+…..x2tanxx=12!loge(2e)2=12loge2+12=12(ln2+1)2 Related content Top 5 Reasons To Study Science After 10th Application for Teaching Job Railway Jobs After 10th CBSE Class 12 Computer Science Answer Key and Paper Analysis 2024 – Student Feedback and Expert View Fractions Class 6 Worksheet Genetics and Evolution | Mind Map Kerala NEET UG Merit List 2023 NEET Exam Centres in Hyderabad 2024 Ecology & Environments NEET Exam Centres in Maharashtra 2024