Q1. \lim_{x\rightarrow 0}\frac{sintanx+ln(\sqrt{1+sin^2x}-sinx)}{ln(1+x^3)}
Q2. A=\begin{bmatrix} 1 &1 &1 \\ -1 &0 &2 \\ 2 &1 &0 \end{bmatrix}, I is the unit matrix of order 3X3 and
aA3 + bA2 + cA + I = B where B is null matrix of order 3X3. then find a,b,c
11An 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{}
66Well 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.
21oh so i got teh right ans [4] tat is 1/2..did jus lik that as done by sir...but ya avik's method also seems fine[12]
106sir, what is wrong in avik's method? then. it also seems fine as does yours
66That'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}}
13Just see my approach...if there's ny fault...
Let tan(x)= t, As x→0, tan(x)→0 => t→0.
This reduces the Expression to...
\lim_{t \rightarrow 0} \frac{sint + ln \frac{\sqrt{1+2t^2}-t}{ \sqrt{1+t^2}}}{ln \{1+ \{tan^{-1}t\}^3\}}
Further...
\lim_{t \rightarrow 0} \frac{sint + ln \{1-t\}}{ ln \{1+ \{tan^{-1}t\}^3\}}
Using Taylor Series Expansion...
\lim_{t \rightarrow 0} \frac{\{t-t^3/3! +t^5/5!+...\}+\{-t+..\}}{\{t-t^3/3+...\}^3+...}
Which is finally giving the answer = -1/6...
106then sir, how do we proceed ? can u just give the steps .. what to replace by what?
1357Also asish
you cannot write (1+x2)-1/2 as 1 (As the denominator is x2)
(1+x2)-1/2=(1-x2/2)
So by your second method we are getting the answer as zero
1357As the denominator is x3
we should better write sinx as x-x3/3!
106sir,
f(x) = ln(\sqrt{1+x^2}-x)
f'(x) = \frac{x/\sqrt{1+x^2}-1}{\sqrt{1+x^2}-x}
f'(x) = \frac{-1}{\sqrt{1+x^2}}
1357I din't get this step .......
so we have lim(x→0) sinx+ln(√(1+x2)-x)/x3
using LH
we get lim(x-->0) [cosx-1/(√(1+x2)]/3x2
106Look 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+x2≈1
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??
106sin(tanx) .. sorry for that
@taran: thanks for that ... I too got that but aakash solutions gave the signs of a,b,c exactly opposite
1|A-λI | =0 IS THE CHARACTERISTIC EQUATION FOR THIS MATRIX
SOLVING THE DETERMINANT WE GET
-λ3 +λ2 +3λ + 1 =0
NOW AS PER CAYLEY HAMILTON THEOREM
EVERY MATRIX SATISFIES ITS CHARACTERISTIC EQUATION
SO WE GET
-A3 +A2 +3A +I =0
HENCE COMPARING COEFEECIENTS WE GET
a = -1
b= 1
c= 3
1someone help in the first one....i hav a method for it but its very lengthy