IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

neil_mahaseth

Please do the first one by taking log

rohitkuruvila

are the answers

1. e^-5

2. 0

rohitkuruvila

are the answers

1. e^-5

2. 0

neil_mahaseth

Please write the method

karthik2007

From there on, it is easy to arrive at the answer

amulye

fr d 1st sum i know a short cut remember d formula

lt f(x)^g(x)=
x--->a
e^lt
x----->a [f(x)-1] g(x)
hence u will get d answer

amulye

rate vme if satisfied

sweet08

i"m getting 2nd ans=1/2

ramkumar_november

l = $\lim_{x\to\ 0}\frac{xtan2x-2xtanx}{(1-cos2x)^2}$

l = $\lim_{x\to\ 0}\frac{x(\frac{2tanx}{1-tan^2x}-2tanx)}{(2sin^2x)^2}$

l = $\lim_{x\to\ 0}\frac{2xtan^3x}{4sin^4x}$

l = $\lim_{x\to\ 0}\frac{1}{2}.\frac{tan^3x}{x^3}.\frac{x^4}{sin^4x}$

l = $\frac{1}{2}$