wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

limx0axaxx

Open in App
Solution

limx0axaxx=limx0(ax1)(ax1)x=limx0ax1xlimx0ax1x=limx0ax1x+limx0ax1(x)=loga+loga=2loge2


flag
Suggest Corrections
thumbs-up
2
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Standard Expansions and Standard Limits
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon