Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-5470-145
+91 7353221155
Use Coupon: CART20 and get 20% off on all online Study Material
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
OTP to be sent to Change
lim x → 0 (x cosx - log (1+x) )/x2 is equal to?
Dear aditi
lim x → 0 (x cosx - log (1+x) )/x2
cosx = 1 - x2/2! +x4 / 4! ............
x cosx = x - x3 /2! + x5/4!..........
log (1+x) = x -x2/2 + x4/2...........
so put the above expansion in the limits we get the limit 1/2
We are all IITians and here to help you in your IIT JEE preparation.
Now you can win by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.
Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar : Click here to download the toolbar..
jitender
simplest way is to use l-hospitals rule.
d^n(numerator)/d^n(denominator) until 0/0 form is removed.then substitute to obtain the limit.
(cosx - xsinx - 1/(1+x))/2x Now since 0/0 form persists again differentiate
(-sinx - sinx -xcosx + 1/(1+x^2))/2 put x=0 to obtain the limit as,1/2
Get your questions answered by the expert for free
You will get reply from our expert in sometime.
We will notify you when Our expert answers your question. To View your Question
Win Gift vouchers upto Rs 500/-
Register Yourself for a FREE Demo Class by Top IITians & Medical Experts Today !