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Limit Problem

\(\displaystyle \lim_{x\to\frac{\pi}{2}} \frac{(1+\tan x)^{10}+(2+\tan x)^{10}+(3+\tan x)^{10}.....(20+\tan x)^{10}}{\tan^{9}x}-20\tan x=?\)

Note by Krishna Jha
3 years, 10 months ago

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Taking the LCM and then considering the expansion of all the terms

Now first 9 terms of expansion will result a polynomial in tanx of degree 8 or lower.

Hence it would tend to zero as x»π/2 Hence writing the last two terms of the expansion

We would get the limit basically as the coefficient of (tan⁡x)^9 in the numerator as =10C9[1+2+3+ …. 20] = 10[20*21/2] = 2100 Ram Prakash Patel Patel · 3 years, 10 months ago

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@Ram Prakash Patel Patel Thanks... It didnt strike me... Krishna Jha · 3 years, 10 months ago

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Where did you get such horrible problem from? Snehal Shekatkar · 3 years, 10 months ago

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@Snehal Shekatkar "Horrible" in what sense???? Btw i got it in a test..could'nt solve it.. Krishna Jha · 3 years, 10 months ago

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@Snehal Shekatkar No way this problem is horrible. Ram Prakash Patel Patel · 3 years, 10 months ago

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@Ram Prakash Patel Patel You mean easy?? Krishna Jha · 3 years, 10 months ago

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take tanx common from each of the terms( i mean those brackets). then take 10^{tanx} common from all the terms and it gets divided with 9^{tanx} in the denominator leaving tanx in the numerator something /tanx tends to 0 because tanx tends to infinity so the inside part will be adding up to 20 making it 20*tanx then it will be zero my answer does not tally with the above one please tell me what is wrong. Santosh Pavan · 3 years, 10 months ago

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