Limit Problem

lim⁡x→π2(1+tan⁡x)10+(2+tan⁡x)10+(3+tan⁡x)10.....(20+tan⁡x)10tan⁡9x−20tan⁡x=?\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=?

#Limits #MathProblem #Math

Note by Krishna Jha
7 years, 8 months ago

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5 votes

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Comments

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 - 7 years, 8 months ago

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Thanks... It didnt strike me...

Krishna Jha - 7 years, 8 months ago

Where did you get such horrible problem from?

Snehal Shekatkar - 7 years, 8 months ago

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"Horrible" in what sense???? Btw i got it in a test..could'nt solve it..

Krishna Jha - 7 years, 8 months ago

No way this problem is horrible.

Ram Prakash Patel Patel - 7 years, 7 months ago

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You mean easy??

Krishna Jha - 7 years, 7 months ago

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 - 7 years, 7 months ago
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