Limits

An alternative method :-

Dividing both numerator and denominator by tanx and using the substitution ln(1+x)\rightarrow x when x tends to 0 we have

\frac{1+\frac{\sqrt{1+sin^2x}}{tanx}-cosx-cotx}{x^2}

The numerator finally becomes 1+cosecx-cosx-cosecx=1-cosx while denominator is x^2

So we require the limit \frac{1-cosx}{x^2} as x tends to 0 which is \frac{1}{2}

\boxed{}

Well Avik has done a lot of mistakes actually[3]
The expression
\lim_{t\to 0}\dfrac{\sin t +\ln\left(\frac{\sqrt{1+2t^2}-t}{\sqrt{1+t^2}}\right)}{\ln(1+(\tan^{-1}x)^3)}
is indeed 1/2.
But the expression
\lim_{t\to 0}\dfrac{\sin t +\ln(1-t)}{\ln(1+(\tan^{-1}x)^3)}
has a limit -∞.
That means the replacement of
\ln\left(\dfrac{\sqrt{1+2t^2}-t}{\sqrt{1+t^2}}\right) with \ln(1-t) is not justified.

Finally the last expression yields -∞ and not -1/6 because he has forgotten the square term that is also present in the expansion.

sir, what is wrong in avik's method? then. it also seems fine as does yours

That's not correct. The required limit is 1/2.
First, apply the L'Hospital rule to get the required limit as
\lim_{x\to 0} \dfrac{\cos(\tan x)\sec^2x -\frac{\cos x}{\sqrt{1+\sin^2x}}}{\frac{3x^2}{1+x^3}}
Applying L'Hospital again does not seem a nice way, since that would lead to unmanageable expressions. Let us try expansions. Write
\cos(\tan x)=1-\frac{\tan^2x}{2}+O(\tan x)
where O(tan x) means higher powers of tan x. Also,
\dfrac{1}{\sqrt{1+\sin^2x}}=1-\dfrac{1}{2}\sin^2x+O(\sin x)
and
\dfrac{1}{1+x^3}=1-x^3+O(x)
Ultimately the higher powers do not contribute anything. So suppressing them, we get the limit as
\lim_{x\to 0} \dfrac{\left(1-\frac{\tan^2x}{2}\right)(1+\tan^2x)-\cos x\left(1-\frac{\sin^2x}{2}\right)}{3x^2(1-x^3)}

\lim_{x\to 0} \dfrac{1+\frac{\tan^2x}{2}-\frac{\tan^4x}{2}-\cos x\left(1-\frac{\sin^2x}{2}\right)}{3x^2(1-x^3)}

Next, we expand sin x, cos x and tan x in their series expansions retaining the first two terms as follows:
\tan x \approx x -\frac{x^3}{3}
\sin x \approx x -\frac{x^3}{6}
\cos x \approx 1 -\frac{x^2}{2}
And the given limit becomes
\lim_{x\to 0} \dfrac{1+\frac{x^2}{2}\left(1-\frac{x^2}{3}\right)^2-\frac{x^4}{2}\left(1-\frac{x^2}{3}\right)^4-\left(1-\frac{x^2}{2}\right)\left(1-\frac{x^2}{2}\left(1-\frac{x^2}{6}\right)^2\right)}{3x^2(1-x^3)}
which further simplifies to
\lim_{x\to 0} \dfrac{1+\frac{x^2}{2}\left(1-\frac{2x^2}{3}\right)-\frac{x^4}{2}\left(1-\frac{4x^2}{3}\right)-\left(1-\frac{x^2}{2}\right)\left(1-\frac{x^2}{2}\left(1-\frac{2x^2}{6}\right)\right)}{3x^2(1-x^3)}
where again binomial expansions have come into picture. Simplification gives
\lim_{x\to 0} \dfrac{\frac{3x^2}{2}-\frac{5x^4}{4}+\frac{3x^6}{4}}{3x^2(1-x^3)}=\boxed{\dfrac{1}{2}}

then sir, how do we proceed ? can u just give the steps .. what to replace by what?

Look at my method.. (there are two giving completely opposite answers)

METHOD 1:
tanx≈sinx≈x
=> sin(tanx) ≈ x
further ln(1+x³) ≈ x³
and the bracket thing simplifies to ln(√1+x²-x)

expanding (1+x)^1/2 binomially,
we get ln(1+12x²-x)

So we get lim(x→0) x+ln(1+12x²-x)x³
which on applying LH gives upon simplification 1/6

METHOD 2:

stop before expanding binomially

so we have lim(x→0) sinx+ln(√1+x²-x)x³

using LH

we get lim(x-->0) cosx-1√1+x²3x²

as x-->0 1+x²â‰ˆ1

so we have lim(x-->0) cosx-13x²
using LH again we get

lim(x-->0) -sinx/6x = -1/6

how can we get 1/6 and -1/6 ... both approaches are correct arent they??

anyone for first one... ?

someone help in the first one....i hav a method for it but its very lengthy