1for first one use this
(1+x)1/x=e^(1x ln (1+x) )
use exp of ln (1+x)
and the apply l hospital
These are some easy ones on limit...
Please explain them and help me finish limits.
Q1) \lim_{x\rightarrow 0} \frac{e-(1+x)^{1/x}}{tan x}
Q2) \lim_{x\rightarrow 0} \left(\frac{(1+x)^{1/x}}{e} \right)^{1/x}
Q3) {P{n}}= \frac{2^{3}-1}{2^{3}+1}.\frac{3^{3}-1}{3^{3}+1}.\frac{4^{3}-1}{4^{3}+1}...\frac{n^{3}-1}{n^{3}+1}
\lim_{n\rightarrow 0} P_{n}=?
Edit: Answer:- 1-e/2 2- e-1/2 3- 2/3
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15 Answers
1
1Yup correct..
Is there any other method...without expansion?
1for second
((1+x)1/x e)1/x = e^((1xln(1+x)-1)1x)
expan ln(1+x) and then apply l hosptital
e^(-1/2)
1hmm...I was trying to avoid expansion.
(Got the answer using expansion)
Still looking for alternative method
Thanx Reimann
1and for third are u sure its summation.....i hav done a similar q but in that each term was multiplied
1why trying to avoid expansion......its good to use them anywhere oder than in boards
1for 3rd one see this
http://targetiit.com/iit-jee-forum/posts/limit-1-12939.html
1I avoid expansion coz I usually mess up at later stages.
106Q1. u can apply L-H rule but it will be very tedious ... btw i never use the exp of (1+x)1/x cant seem to remember it .. so i use LH only