L'Hopital Isn't The Saviour This Time!

Calculus Level 4

lim x 0 1 k = 1 23 cos k x x 2 = ? \large \lim_{x \to 0} \frac{ 1- \displaystyle \prod_{k=1}^{23} \cos kx}{x^2} = \, ?


The answer is 2162.

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8 solutions

Michael Mendrin
Jan 14, 2017

Since the limit is x 0 x \rightarrow 0 , we'll use the series expansion of the Cosine function, so that we have

lim x 0 1 x 2 ( 1 k = 1 23 ( 1 k 2 x 2 2 ! + A x B + 2 ) ) \displaystyle \lim_{x \to 0} \dfrac { 1 }{ { x }^{ 2 } } \left( 1-\displaystyle\prod _{ k=1 }^{ 23 }{ \left( 1-\dfrac { { k }^{ 2 }{ x }^{ 2 } }{ 2! } +A{ x }^{ B+2 } \right) } \right)

lim x 0 1 x 2 ( 1 ( 1 x 2 2 ! k = 1 23 k 2 + A x B + 2 ) ) \displaystyle \lim_{x \to 0} \dfrac { 1 }{ { x }^{ 2 } } \left( 1-\left( 1-\dfrac { { x }^{ 2 } }{ 2! } \displaystyle\sum _{ k=1 }^{ 23 }{ { k }^{ 2 }+A{ x }^{ B+2 } } \right) \right)

lim x 0 1 x 2 ( 1 ( 1 x 2 2 ! 4324 + A x B + 2 ) ) \displaystyle \lim_{x \to 0} \dfrac { 1 }{ { x }^{ 2 } } \left( 1-\left( 1-\dfrac { { x }^{ 2 } }{ 2! } 4324+A{ x }^{ B+2 } \right) \right)

so that the limit value is 2162 2162

A A and B > 0 B>0 stand for arbitrary nonzero values.

wow that was quick

space sizzlers - 4 years, 4 months ago

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Yes, usually for such limit problems, considering the (Laurent) polynomial expansion makes quick work. That's the generalized approach of L'Hopital.

Calvin Lin Staff - 4 years, 4 months ago

Let's generalize this for n:

= lim x 0 1 k = 1 n cos k x x 2 =\large \displaystyle \lim_{x \to 0} \frac{\large \displaystyle 1- \prod_{k=1}^{n} \cos kx}{\large \displaystyle x^2}

lim x 0 1 cos x cos 2 x cos 3 x . . . . . cos n x x 2 \large \displaystyle\lim_{x \to 0} \frac{1-\cos x\cdot \cos 2x\cdot \cos 3x\cdot .....\cdot \cos nx}{x^2}

= lim x 0 1 cos x + cos x cos x cos 2 x cos 3 x . . . . . cos n x x 2 =\large \displaystyle\lim_{x \to 0} \frac{1-\cos x+\cos x-\cos x\cdot \cos 2x\cdot \cos 3x\cdot .....\cdot \cos nx}{x^2}

= lim x 0 1 cos x x 2 + lim x 0 cos x lim x 0 ( 1 cos 2 x cos 3 x . . . cos n x ) x 2 =\large \displaystyle\lim_{x \to 0} \frac{1-\cos x}{x^2}+\lim_{x \to 0}\cos x \lim_{x \to 0} \frac{\left(1-\cos 2x\cdot \cos 3x\cdot ...\cdot \cos nx\right)}{x^2}

= 1 2 2 + lim x 0 ( 1 cos 2 x + cos 2 x cos 2 x cos 3 x . . . cos n x ) x 2 =\large \displaystyle \frac{1^2}{2}+ \lim_{x \to 0} \frac{\left(1-\cos 2x+\cos 2x-\cos 2x\cdot \cos 3x\cdot ...\cdot \cos nx\right)}{x^2}

= 1 2 2 + lim x 0 1 cos 2 x x 2 + lim x 0 cos 2 x lim x 0 ( 1 cos 3 x cos 4 x . . . cos n x ) x 2 =\large \displaystyle \frac{1^2}{2}+ \lim_{x \to 0} \frac{1-\cos 2x}{x^2}+\lim_{x \to 0}\cos 2x \lim_{x \to 0} \frac{\left(1-\cos 3x\cdot \cos 4x\cdot ...\cdot \cos nx\right)}{x^2}

= 1 2 2 + 2 2 2 + lim x 0 ( 1 cos 3 x cos 4 x . . . cos n x ) x 2 =\large \displaystyle \frac{1^2}{2}+ \frac{2^2}{2} + \lim_{x \to 0} \frac{\left(1-\cos 3x\cdot \cos 4x\cdot ...\cdot \cos nx\right)}{x^2}

= 1 2 2 + 2 2 2 + 3 2 2 + . . . . . . . + ( n 1 ) 2 2 + lim x 0 1 cos n x x 2 =\large \displaystyle \frac{1^2}{2}+ \frac{2^2}{2} + \frac{3^2}{2} + ....... + \frac{(n-1)^2}{2} + \lim_{x \to 0} \frac{1-\cos nx}{x^2}

= 1 2 2 + 2 2 2 + 3 2 2 + . . . . . . . + ( n 1 ) 2 2 + n 2 2 =\large \displaystyle \frac{1^2}{2}+ \frac{2^2}{2} + \frac{3^2}{2} + ....... + \frac{(n-1)^2}{2} + \frac{n^2}{2}

= 1 2 + 2 2 + 3 2 + . . . + ( n 1 ) 2 + n 2 2 = k = 1 n k 2 2 = n ( n + 1 ) ( 2 n + 1 ) 12 =\large \displaystyle \frac{1^2 + 2^2 + 3^2 + ... + (n-1)^2 + n^2}{2} = \frac{\large \displaystyle \sum_{k=1}^{n} k^2}{2}=\boxed{\frac{n(n+1)(2n+1)}{12}}

For our question, substituting n = 23 n=23 gives 23 24 47 12 = 2162 \frac{23 \cdot 24 \cdot 47}{12} = 2162

A simpler way to present it is via induction (to replace the repeated factorization that you have).
Using your idea of taking out a single term, we could modify it to

lim x 0 1 k = 1 n cos k x x 2 = lim x 0 1 cos n x x 2 + cos n x ( 1 k = 1 n 1 cos k x ) x 2 = n 2 2 + 1 × k = 1 n 1 k 2 2 \lim_{x \rightarrow 0 } \frac{ 1 - \prod_{k=1}^n \cos kx } { x^2 } =\lim_{x \rightarrow 0 } \frac{ 1 - \cos nx } { x^2 } + \frac{ \cos nx \left( 1 - \prod_{k=1}^{n-1} \cos kx \right)} {x^2 } = \frac{ n^2 } { 2 } + \frac{ 1 \times \sum_{k=1}^{n-1} k^2 } { 2 }

Calvin Lin Staff - 4 years, 4 months ago

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I guess this is better. Thanks :)

Arkajyoti Banerjee - 4 years, 4 months ago

Interesting solution and a nice generalization!

Christopher Boo - 4 years, 4 months ago

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Thank you :D

Actually I saw this question in a competitive exam's paper:

lim x 0 1 cos x cos 2 x cos 3 x ( sin x ) 2 \lim_{x\to 0} \frac{1-\cos x \cos 2x \cos 3x}{(\sin x)^2}

So I thought of generalizing this for n.

Arkajyoti Banerjee - 4 years, 4 months ago

We usually do not expect a limit problem to yield a big answer.

Kartik Sharma - 4 years, 4 months ago

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Try expecting the unexpected xD

Arkajyoti Banerjee - 4 years, 4 months ago
Ivan Koswara
Jan 20, 2017

L'Hopital is not the savior? Think again.

Substitute x = 0 x = 0 to the numerator and the denominator. Both result in 0, so we have the indeterminate form 0 0 \frac{0}{0} . By L'Hopital's theorem, we can differentiate both the numerator and the denominator, and the limit will stay equal.

The denominator becomes 2 x 2x . In the numerator, the 1 1 vanishes. The remaining term, cos k x - \prod \cos kx , splits into 23 terms by product rule, once for differentiating cos k x \cos kx for each k k . It becomes k = 1 23 k cos x cos 2 x cos ( k 1 ) x sin k x cos ( k + 1 ) x cos 23 x \sum_{k=1}^{23} k \cdot \cos x \cdot \cos 2x \cdot \ldots \cdot \cos (k-1)x \cdot \sin kx \cdot \cos (k+1)x \cdot \ldots \cdot \cos 23x ; the factor cos k x \cos kx is differentiated into k sin k x -k \sin kx , and the negative sign cancels.

Now, try substituting again. The numerator is still 0 because each term is zero due to the sin k x \sin kx term. The denominator is also still 0. So use L'Hopital's theorem again; differentiate again.

The denominator becomes simply 2. In the numerator, each term of the summation splits into 23 terms by product rule. One term comes from differentiating sin k x \sin kx ; this results in k 2 i = 1 23 cos i x k^2 \cdot \prod_{i=1}^{23} \cos ix (the term that contains sin k x \sin kx also contains the factor k k , since we differentiated cos k x \cos kx into k sin k x -k \sin kx before). The other comes from differentiating some cos i x \cos ix , which leaves the sin k x \sin kx factor intact.

Finally, substitute again. In the numerator, every term that contains sin k x \sin kx is gone, leaving only the terms in the form k 2 cos i x k^2 \cdot \prod \cos ix . In all these terms, the cosines evaluate to 1, so all the terms are in the form k 2 k^2 . So the numerator is k = 1 23 k 2 = 23 24 47 6 \sum_{k=1}^{23} k^2 = \frac{23 \cdot 24 \cdot 47}{6} . The denominator is simply 2. So substituting x = 0 x = 0 to the expression gives 23 24 47 6 2 = 2162 \frac{23 \cdot 24 \cdot 47}{6 \cdot 2} = \boxed{2162} , which must be the answer because the numerator and the denominator are both continuous at x = 0 x = 0 .


This method can be generalized to evaluate

lim x 0 1 k = 1 n cos k x x m \lim_{x \to 0} \dfrac{1 - \prod_{k=1}^n \cos kx}{x^m}

for any positive integers n , m n,m , although the solution will become more messy. (But at least we get the denominator generalized too, compared to other solutions that only do m = 2 m = 2 !) If m m is odd, the answer is 0 (there's an odd number of sines remaining, in particular at least one); otherwise, each surviving term in the final summation is in the form coefficient ( 1 ) m / 2 + 1 k = 1 n cos k x \text{coefficient} \cdot (-1)^{m/2 + 1} \cdot \prod_{k=1}^n \cos kx , where the coefficient comes from differentiating several of the cosine factors. Each factor must be differentiated an even number of times (to turn them into cosines again). We just need to count the number of ways to get the same coefficient.

For example, for m = 6 m = 6 , there are three ways to get all cosines:

  • A single factor cos a x \cos ax is differentiated 6 times. The coefficient is a 6 a^6 , and this can be achieved in one way.
  • A factor cos a x \cos ax is differentiated 4 times, and another factor cos b x \cos bx is differentiated 2 times. The coefficient is a 4 b 2 a^4 b^2 , and this can be achieved in ( 6 2 ) = 15 \binom{6}{2} = 15 ways.
  • Three factors cos a x , cos b x , cos c x \cos ax, \cos bx, \cos cx are differentiated 2 times each. The coefficient is a 2 b 2 c 2 a^2 b^2 c^2 , and this can be achieved in ( 6 2 , 2 , 2 ) = 90 \binom{6}{2, 2, 2} = 90 ways. However, permuting a , b , c a, b, c will give the same coefficient, so we count each coefficient 3 ! = 6 3! = 6 times. The effective multiplier is 90 6 = 15 \frac{90}{6} = 15 .

Thus summing over everything, the final coefficient will be

a a 6 + 15 a b a 4 b 2 + 15 $a, b, c$ distinct a 2 b 2 c 2 \sum_{a} a^6 + 15 \cdot \sum_{a \neq b} a^4 b^2 + 15 \cdot \sum_{\text{\$a, b, c\$ distinct}} a^2 b^2 c^2

And how do we compute it? Well, add 14 a 6 14 a 6 14 \sum a^6 - 14 \sum a^6 , and we can gather those with multiplier 15 as simply ( a 2 ) 3 (\sum a^2)^3 . Thus the answer is

15 ( a = 1 n a 2 ) 3 14 a = 1 n a 6 15 \cdot \left( \sum_{a=1}^n a^2 \right)^3 - 14 \cdot \sum_{a=1}^n a^6

which can be easily computed by Newton's identities.

The title meant that you cannot solve this by directly applying L'Hopital's Rule to the expression. Besides, this question can be solved using a rather shorter method.

Arkajyoti Banerjee - 4 years, 4 months ago

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But I did apply L'Hopital's rule directly to the expression. You can actually apply it twice and you immediately get the answer. The numerator will become very long, but it is still purely L'Hopital's rule.

Example where there are three cosines, spelled out fully in L'Hopital's rule:

lim x 0 1 cos x cos 2 x cos 3 x x 2 = lim x 0 sin x cos 2 x cos 3 x + 2 cos x sin 2 x cos 3 x + 3 cos x cos 2 x sin 3 x 2 x = lim x 0 cos x cos 2 x cos 3 x 2 sin x sin 2 x cos 3 x 3 sin x cos 2 x sin 3 x 2 sin x sin 2 x cos 3 x + 4 cos x cos 2 x cos 3 x 6 cos x sin 2 x sin 3 x 3 sin x cos 2 x sin 3 x 6 cos x sin 2 x sin 3 x + 9 cos x cos 2 x cos 3 x 2 = 1 1 1 1 2 0 0 1 3 0 1 0 2 0 0 1 + 4 1 1 1 6 1 0 0 3 0 1 0 6 1 0 0 + 9 1 1 1 2 = 1 0 0 0 + 4 0 0 0 + 9 2 = 14 2 = 7 \begin{aligned} \displaystyle \lim_{x \to 0} \frac{1 - \cos x \cos 2x \cos 3x}{x^2} &= \lim_{x \to 0} \frac{\sin x \cos 2x \cos 3x + 2 \cos x \sin 2x \cos 3x + 3 \cos x \cos 2x \sin 3x}{2x} \\ &= \lim_{x \to 0} \frac{\cos x \cos 2x \cos 3x - 2 \sin x \sin 2x \cos 3x - 3 \sin x \cos 2x \sin 3x - 2 \sin x \sin 2x \cos 3x + 4 \cos x \cos 2x \cos 3x - 6 \cos x \sin 2x \sin 3x - 3 \sin x \cos 2x \sin 3x - 6 \cos x \sin 2x \sin 3x + 9 \cos x \cos 2x \cos 3x}{2} \\ &= \frac{1 \cdot 1 \cdot 1 \cdot 1 - 2 \cdot 0 \cdot 0 \cdot 1 - 3 \cdot 0 \cdot 1 \cdot 0 - 2 \cdot 0 \cdot 0 \cdot 1 + 4 \cdot 1 \cdot 1 \cdot 1 - 6 \cdot 1 \cdot 0 \cdot 0 - 3 \cdot 0 \cdot 1 \cdot 0 - 6 \cdot 1 \cdot 0 \cdot 0 + 9 \cdot 1 \cdot 1 \cdot 1}{2} \\ &= \frac{1 - 0 - 0 - 0 + 4 - 0 - 0 - 0 + 9}{2} \\ &= \frac{14}{2} \\ &= 7 \end{aligned}

First and second lines are L'Hopital's rule; third line is substitution due to continuity; the rest is computation. Standard application(s) of L'Hopital's rule. It's easy to generalize that to the actual problem of 23 cosines, if you want to spell out everything.

Ivan Koswara - 4 years, 4 months ago

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@Arkajyoti Banerjee I agree with Ivan. Basically, you're trying to find the first non-zero coefficient in the (Laurent) polynomial expansion of these terms. Both L'hopital's rule (Ivan's solution) and converting to Taylor polynomials (Michael's solution) are the same way to do this.

In particular, because the constant and the linear term are 0 (as in Michael's polynomials), this implies that we have to apply L'hopital twice. Once to remove the constant, once to remove the linear term. Of course, the expansion might be tedious (as Ivan pointed out), which is why I prefer to transition directly to the Taylor polynomial forms for complicated terms.

Calvin Lin Staff - 4 years, 4 months ago

I just meant that this problem cannot be solved by applying L'Hopital's Rule without algebraic manipulation. You just applied it for the expression till 3 terms and followed the pattern. The original expression contained 23. (Generalisation contained n). So you don't call this a direct application, do you?

Arkajyoti Banerjee - 4 years, 4 months ago

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@Arkajyoti Banerjee I can directly apply it (twice) to the expression; the method is exactly the same. It will simply be tedious (see my example with n = 3 n = 3 above, and imagine it extending to n = 23 n = 23 ; a single line will take an entire page of paper) and not particularly illuminating (generalization is always good in math since it gives insight; instead of "why 2162" you get "oh, because (\frac{1^2 + 2^2 + \ldots + n^2}{2}").

Ivan Koswara - 4 years, 4 months ago

Yes, we can directly apply, I say..

Think Bhai Think...

Differentiate and then putting sin x x , 2 sin 2 x 2 x , . . . = 1 \dfrac{\sin x}{x} , 2\dfrac{\sin2x}{2x}, ...= 1 , We get it .

I got it like that

Md Zuhair - 4 years, 2 months ago
Chew-Seong Cheong
Jan 15, 2017

Relevant wiki: Maclaurin Series

L = lim x 0 1 k = 1 23 cos k x x 2 By Maclaurin series = lim x 0 1 x 2 ( 1 k = 1 23 ( 1 x 2 2 ! + x 4 4 ! + ) ( 1 2 2 x 2 2 ! + 2 4 x 4 4 ! + ) ( 1 3 2 x 2 2 ! + 3 4 x 4 4 ! + ) ( 1 2 3 2 x 2 2 ! + 2 3 4 x 4 4 ! + ) ) = lim x 0 1 x 2 ( 1 ( 1 k = 1 23 k 2 x 2 2 ! + O ( x 4 ) ) ) = lim x 0 1 x 2 ( ( 23 ) ( 24 ) ( 46 + 1 ) x 2 6 2 ! O ( x 4 ) ) = lim x 0 1 x 2 ( 2162 x 2 O ( x 4 ) ) = lim x 0 ( 2162 O ( x 2 ) ) = 2162 \begin{aligned} L & = \lim_{x \to 0} \frac {1-\prod_{k=1}^{23} {\color{#3D99F6}\cos {kx}}}{x^2} \quad \quad \small \color{#3D99F6} \text{By Maclaurin series} \\ & = \small \lim_{x \to 0} \frac 1{x^2} \left(1 - \prod_{k=1}^{23} \left(1-\frac {x^2}{2!} + \frac {x^4}{4!} + \cdots \right) \left(1-\frac {2^2x^2}{2!} + \frac {2^4x^4}{4!} + \cdots \right) \left(1-\frac {3^2x^2}{2!} + \frac {3^4x^4}{4!} + \cdots \right) \cdots \left(1-\frac {23^2x^2}{2!} + \frac {23^4x^4}{4!} + \cdots \right) \right) \\ & = \lim_{x \to 0} \frac 1{x^2} \left(1- \left(1 - \frac {\sum_{k=1}^{23} k^2 x^2}{2!} + O(x^4) \right) \right) \\ & = \lim_{x \to 0} \frac 1{x^2} \left( \frac {(23)(24)(46+1) x^2}{6 \cdot 2!} - O(x^4) \right) \\ & = \lim_{x \to 0} \frac 1{x^2} \left(2162x^2 - O(x^4) \right) \\ & = \lim_{x \to 0} \left(2162 - O(x^2) \right) \\ & = \boxed{2162} \end{aligned}

I was expecting a solution from you for this one, and it is breathtakingly awesome as always! :D

Arkajyoti Banerjee - 4 years, 4 months ago

A solution using probabilistic method.

The first key observation is the identity k = 1 n cos ( k θ ) = 1 2 n e i { 1 , + 1 } , 1 i n cos ( ( e 1 + 2 e 2 + + n e n ) θ ) \prod_{k=1}^n \cos(k \theta)=\frac{1}{2^n}\sum_{e_i\in \{-1,+1\},\\1\le i\le n}\cos((e_1+2e_2+\cdots+ne_n)\theta) this can be proved using the Euler's formula and induction. Next, define the random variables X i , 1 i n X_i,1\le i\le n , such that P ( X i = 1 ) = P ( X i = + 1 ) = 1 2 P(X_i=-1)=P(X_i=+1)=\frac{1}{2} , and let X i X_i are all independent. Further define the vectors X = [ X 1 X 2 X n ] T , a = [ 1 2 n ] T \mathbf{X}=[X_1\ X_2\ \cdots\ X_n]^T,\ \mathbf{a}=[1\ 2\ \cdots\ n]^T . Then, we get k = 1 n cos ( k θ ) = E ( cos ( a T X θ ) ) \prod_{k=1}^n \cos(k \theta)=E(\cos(\mathbf{a}^T\mathbf{X}\theta)) where the expectation is taken with respect to X i , 1 i n X_i,\ 1\le i\le n . Then, the given limit can be evaluated as lim x 0 1 E ( cos ( a T X x ) ) x 2 = lim x 0 E [ ( 2 sin 2 ( a T X x / 2 ) ) x 2 ] = 1 2 E [ ( a T X ) 2 ] \lim_{x\to 0}\frac{1-E(\cos(\mathbf{a}^T\mathbf{X}x))}{x^2}=\lim_{x\to 0}E\left[\frac{(2\sin^2(\mathbf{a}^T\mathbf{X}x/2))}{x^2}\right]\\=\frac{1}{2}E[(\mathbf{a}^T\mathbf{X})^2] where the last step follows from Monotone Convergence Theorem . Now, E [ ( a T X ) 2 ] = a T E ( X X T ) a E[(\mathbf{a}^T\mathbf{X})^2]=\mathbf{a}^TE(\mathbf{X}\mathbf{X}^T)\mathbf{a} Note that that the ( i , j ) (i,j) th entry of the matrix E ( X X T ) E(\mathbf{X}\mathbf{X}^T) is E ( X i X j ) = E ( X i ) E ( X j ) = 0 E(X_iX_j)=E(X_i)E(X_j)=0 , whenever i j i\ne j , because of independence and the fact that the random variables are zero mean by definition. The diagonal entries of the matrix are E ( X i 2 ) = 1 E(X_i^2)=1 . Hence, the matrix E ( X X T ) E(\mathbf{X}\mathbf{X}^T) is the identity matrix of order n n . Thus, the limit evaluates to a T a / 2 = i = 1 n a i 2 2 = i = 1 n i 2 2 = n ( n + 1 ) ( 2 n + 1 ) 12 \mathbf{a}^T\mathbf{a}/2=\frac{\sum_{i=1}^n a_i^2}{2}=\frac{\sum_{i=1}^n i^2}{2}=\frac{n(n+1)(2n+1)}{12} , which gives the answer to the given problem, for n = 23 n=23 , as 2162 \boxed{2162} .

More generally, using the same approach as above, we can evaluate the limit lim x 0 1 k = 1 n cos ( θ k x ) x 2 \lim_{x\to 0}\frac{1-\prod_{k=1}^n \cos(\theta_k x)}{x^2} to be ( k = 1 n θ k 2 ) / 2 (\sum_{k=1}^n \theta_k^2)/2 .

I'm an eleventh standard student. But I will look this up for sure lol.

Arkajyoti Banerjee - 4 years, 4 months ago

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The expectation that I have used is not anything advanced, and is mostly used to represent things in a more compact way. I am sure you can understand the proof once you know what expectation is, try reading the basic of expectation from here .

Samrat Mukhopadhyay - 4 years, 4 months ago

Wow, that's really interesting!

Calvin Lin Staff - 4 years, 4 months ago

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Thanks Calvin! I think this method can be used in more general problems where some kind of product structure is present.

Samrat Mukhopadhyay - 4 years, 4 months ago
Rohan Shrothrium
Jan 21, 2017

Frankly speaking L'Hopital was the saviour. :P Just diffrentiate the numarator....... you get the limit as i = 1 n \displaystyle \sum_{i=1}^n ( i ) 2 2 \frac {(i)^2}2

Can you be kind enough to elaborate your solution?

Arkajyoti Banerjee - 4 years, 4 months ago
Kushal Bose
Jan 14, 2017

Let f ( x ) = c o s x . c o s 2 x . . . . c o s 23 x f ( x ) = s i n x . cos 2 x . c o s 3 x . . . c o s 23 x 2 c o s x . s i n 2 x . c o s 3 x . . . . . c o s 23 x . . . . 23 c o s x . c o s 2 x . . . . s i n 23 x f ( x ) = c o s x . c o s 2 x . . . c o s 23 x . t a n x 2 c o s x . c o s 2 x . . . c o s 23 x . t a n 2 x . . . . . 23 c o s x . c o s 2 x . . . . c o s 23 x . t a n 23 x f ( x ) = f ( x ) ( t a n x + 2 t a n 2 x + . . . . . . . + 23 t a n 23 x ) f ( x ) = f ( x ) ( t a n x + 2 t a n 2 x + . . . . . . . + 23 t a n 23 x ) f ( x ) ( s e c 2 x + 2 2 s e c 2 2 x + 3 2 s e c 2 3 x + . . . . . + 2 3 2 s e c 2 23 x ) f(x)=cosx.cos2x....cos23x \\ f'(x)=-sinx.\cos2x.cos3x...cos 23x -2 cosx.sin 2x.cos3x.....cos 23x -....- 23cosx.cos2x....sin 23x \\ f'(x)=-cosx.cos2x...cos23x.tanx - 2cosx.cos2x...cos23x.tan2x -.....- 23 cosx.cos2x....cos23x.tan23x \\ f'(x)=-f(x)(tanx+2tan2x+.......+23 tan23x) \\ f''(x)=-f'(x)(tanx+2tan2x+.......+23 tan23x) - f(x)(sec^2x + 2^2 sec^2{2x}+ 3^2 sec^2{3x} +.....+ 23^2 sec^2{23x})

Now come to th limit

lim x 0 1 f ( x ) x 2 = lim x 0 f ( x ) 2 x . . . . . . . . . . . . . . . . . . . . . . . . . . . . As 0/0 form = lim x t o 0 f ( x ) 2 . . . . . . . . . . . . . . . . . . . . As f’(0)=0 and also 0/0 form = f ( 0 ) 2 = 1 2 + 2 2 + 3 2 + . . . . . . . . . . + 2 3 2 2 = 2162 \lim_{x \to 0} \dfrac{1-f(x)}{x^2} \\ =\lim_{x \to 0} \dfrac{-f'(x)}{2x}............................\text{As 0/0 form} \\ =\lim_{x to 0} \dfrac{-f''(x)}{2}....................\text{As f'(0)=0 and also 0/0 form} \\ =\dfrac{-f''(0)}{2}=\dfrac{1^2+2^2+3^2+..........+23^2}{2}=2162

f ( 0 ) = 1 ; f ( 0 ) = f ( 0 ) × 0 ( 1 2 + 2 2 + 3 2 + . . . . . + 2 3 2 ) f(0)=1 ; f''(0)=f'(0) \times 0 - (1^2+2^2+3^2+.....+23^2)

@Kushal Bose - Great Solution and indeed a good method for differentiation for the function. Solved in exactly the same manner

Anubhav Tyagi - 4 years, 4 months ago

Same way did it

Md Zuhair - 4 years, 2 months ago
Raunak Agrawal
Feb 27, 2017

I actually did it with L- Hopital only . Lol

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