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Differential Calculus

Blazing goIITian

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5 Sep 2008 16:32:45 IST

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747

Limits (some easy qns )

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$1)lim_{x\to infinity} (x - ln \cosh x) \ where \ \cosh t = \frac{e^t +e^{-t}}{2} \\ \\ 2)Let \ P_n = a^{P_{n-1}} - 1, \ where \ n=2,3,.... \ and \\ \\ Let \ P_1 =a^x -1 \ where \ a \ belongs \ to \ R^{-} \ then \ evaluate \\ \\ \ lim_{x\to 0} \frac{P_n}{x} \\ \\ 3)If \ s_n \ be \mbox{the sum of n terms of the series }-> \\ \\ \sin x + \sin 2x + \sin 3x + ... + \sin nx , \ then \ show \ that \\ \\ lim_{n\to infinity} \frac{s_1 + s_2 - .... + s_n}{n}=\frac{\cot \frac{x}{2}}{2} \ (x !=2k \pi ,k \ belongs \ to \ I) \\ \\ 4)\mbox{A circular arc of radius 1 subtends an angle of x radians, } 0<x<pi/2 \ as \ shown\\ \\ \mbox{The point C is the intersection of the two tangent lines at A and B .} \\ \\\mbox{Let T(x) be the area of triangle ABC and let S(x) be the area of the shaded region } \\ \\ Compute -> lim_{x\to 0} \frac{T(x)}{S(x)}\\ \\ 5)\mbox{Through a point A on a circle , a chord AP is drawn and on the tangent at A a point T is taken such that }AT=AP. \\ \\ \mbox{If TP produced meets the diameter through A at Q , prove that the limiting}\\ \\ \mbox{ value of AQ when P moves upto A is double the diameter of the circle }$

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Comments (4)

anchit saini

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5 Sep 2008 16:38:26 IST

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In Q4) , its l(ell) radians

Anand Hegde

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5 Sep 2008 20:25:36 IST

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Posting after a long time :D

ans for the fourth qn- 3/2 ?

Hari Shankar

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6 Sep 2008 09:46:24 IST

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1st one:

$\lim_{x \rightarrow \infty} x - \ln \left(\frac{e^x + e^{-x}}{2} \right) = \lim_{x \rightarrow \infty} \ln e^x - \ln \left(\frac{e^x + e^{-x}}{2} \right) \\ \\ = \lim_{x \rightarrow \infty}\ln \left(\frac{2 e^x}{e^x + e^{-x}}\right) = \lim_{x \rightarrow \infty}\ln \left(\frac{2 }{1 + e^{-2x}}\right) = \ln 2$

Hari Shankar

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6 Sep 2008 10:32:26 IST

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2nd is $\ln^n a$ with the reasonbale assumption you meant to write $a \in \mathbb{R^+}$

This is best proved by induction. Assume the result for n = k

Now,

$\lim_{x \rightarrow 0} \frac{P_{k+1}}{x} = \lim_{x \rightarrow 0} \frac{a^{P_k} - 1}{x} = \lim_{x \rightarrow 0} \frac{a^{P_k} - 1} {a^{P_{k-1}} -1} \frac{a^{P_{k-1}} -1}{x} \\ \\ = \lim_{x \rightarrow 0} \frac{a^{P_k} - 1} {a^{P_{k-1}} -1} \lim_{x \rightarrow 0}\frac{a^{P_{k-1}} -1}{x} \ \text{(with the assumption that the limits involved are finite)} \\ \\ = \lim_{x \rightarrow 0} \frac{a^{P_k} - 1} {a^{P_{k-1}} -1} \lim_{x \rightarrow 0} \frac{P_k}{x} = \lim_{x \rightarrow 0} \frac{a^{P_k} - 1} {a^{P_{k-1}} -1} \ln^k a$ Now,

The last step uses the induction assumption that for n = k, the limit is ln^ka

Now the next step requires us to use $\lim_{x \rightarrow 0} a^{P_{k-1}} - 1 = 0$ which is left as an exercise to the reader (expert's privileges)

$\because \ P_k = a^{P_{k-1}}-1 \\ \\ \lim_{x \rightarrow 0} \frac{a^{P_k} -1}{a^{P_{k-1}} -1} =\lim_{x \rightarrow 0} \frac{a^{a^{P_{k-1}}-1}-1} {a^{P_{k-1}} -1} = \lim_{y \rightarrow 0} \frac{a^y-1}{y} (\text {where} \ y = a^{P_{k-1}} -1)\\ \\ = \ln a $

So, now we can write for n = k+1 the limit is ln^k+1a

Verifying for k =1, we thus prove that it is true for all n

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