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Class 12 Tuition > Learn Class XI-XII Tuition (PUC) > lim(x ?0)? (1+sin?x )^cot? x = ?
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lim(x ?0)? (1+sin?x )^cot? x = ?

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SME in Mathematics & Statistics

Dear Akshay, the question is not clear as there is some typing error. Please post the question again.
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lim(x tends to 0) (1+sinx )^cot x = ?
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The question can be stated as follows: To find: lim x--->0 (1+(sin x)^2)^ Solution: Step 1 (cot x)^2 = 1+ (cosec x)^2 Step 2 putting this identity in the limit we get, lim x--->0 (1+(sin x)^2)^ Step 3 Open the exponent. lim x--->0 * = lim x--->0 * lim x--->0 =1...
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The question can be stated as follows: To find: lim x--->0 (1+(sin x)^2)^ [(cot x)^2] Solution: Step 1 (cot x)^2 = 1+ (cosec x)^2 Step 2 putting this identity in the limit we get, lim x--->0 (1+(sin x)^2)^ [1+ (cosec x)^2] Step 3 Open the exponent. lim x--->0 [(1+(sin x)^2)^1]* [(1+(sin x)^2)^(cosec x)^2] = lim x--->0[(1+(sin x)^2)^1] * lim x--->0 [(1+(sinx)^2)^(cosec x)^2] =1 * lim x--->0 [(1+(sin x)^2)^(cosec x)^2] = lim x--->0 [(1+(sin x)^2)^(cosec x)^2] Step 4 let (sin x)^2 =t then (cosec x)^2 = 1/t as x tends to 0, t also tends to 0=lim t--->0 (1+t)^(1/t) = e Therefore, lim x--->0 (1+(sinx)^2)^ [(cotx)^2] = e read less
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Mathematics for JEE Mains/Advanced, XI & XII (All Boards)

Hi Akshay, Answer to this is "e". It is of 1 power inf form.
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Mr Kumar, The Question you have entered is not clear. reenter the text properly so that we can understand and answer.
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this is indefinite form of 1^infinity. using the result lim(x -> 0) (1+x)^(1/x) = e. this can be written as lim (x->0) (1 + sin x)^( (1/ sin x) * cos x) = e^ cos 0=e. Also we can use L'hospital's rele to solve this. Simply solve lim (x->0) cot x ln (1 + sin x) = lim (x ->0) (ln (1 + sin x))/(tan...
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this is indefinite form of 1^infinity. using the result lim(x -> 0) (1+x)^(1/x) = e. this can be written as lim (x->0) (1 + sin x)^( (1/ sin x) * cos x) = e^ cos 0=e. Also we can use L'hospital's rele to solve this. Simply solve lim (x->0) cot x ln (1 + sin x) = lim (x ->0) (ln (1 + sin x))/(tan x). In this case also answer is e. read less
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