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Grade: 11

                        

Please solve it as soon as possible. ________________________________________________________________________________________________

5 months ago

Answers : (2)

Arun
24742 Points
							
Dear student
 
image is not clear please check entry post the question with a clear image I will be happy to help you
5 months ago
Aditya Gupta
2065 Points
							
dear student, arun is just making excuses. trust me the pic is clearly visible. he cn also download it for better visibility.
coming to ur ques, such ques are best done using series expansion.
f(x)= x(x^1/1! – x^3/3! + x^5/5! – …...)/2 – 1 + 1 – x^2/2! + x^4/4! – ….....
= x^2/2 – x^4/12 + …... – x^2/2! + x^4/24 – …......
= – x^4/12 + x^6/240 – ….. + x^4/24 – …
= x^4( – 1/12+x^2/240 – …. + 1/24 – …...)
hence, Lt f(x)/x^k= Lt [x^4/x^k]*( – 1/12+x^2/240 – …. + 1/24 – …...)
= Lt [x^4/x^k]* Lt ( – 1/12+x^2/240 – …. + 1/24 – …...)
= Lt x^(4 – k) * ( – 1/12 + 1/24)
now, Lt x^(4 – k) exists finitely and non zero only when k=4. if k is less than 4, then 4 – k is greater than zero and hence lim becomes 0. if k is greater than 4, then k – 4 is greater than zero and hence lim becomes Lt 1/x^(k – 4)= ± infinity
so, k=4
KINDLY APPROVE :))
5 months ago
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