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

Limits that lead to the indeterminate forms 1,00,0 can sometimes be solved taking logarithm first and then using L' Hospital's rule

Let limxa(f(x))g(x) is in the form of 0, it can be written as elimxag(x)lnf(x)=eL where L=limxalnf(x)1g(x) is form and can be solved using L' Hopital's rule.

The value of limx[(lnx)12x+x1xn] nN is :

A
2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
e12
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
e
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 2
limx[(lnx)12x+x1xn]=limx(lnx)12x+limxx1xn(1)

Let,
m=limx(lnx)12x&n=x1xn
Taking log on both sides on both equations,
logm=limxlog(lnx)2x and logn=limxlogxxn
Clearly, both the limits are of form , and applying L'hopital's rule,
logm=limx1x2lnx and logn=limx1xnxn1
logm=0 &logn=0
m=1 &n=1
Putting in (1), we get,
limx[(lnx)12x+x1xn]=limx(lnx)12x+limxx1xn=1+1=2

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Algebra of Complex Numbers
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon