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]
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
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19 Answers
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.
21
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