Lt (x-0) xcosx – log(1+x) x 2 how to solve these kind of questions without using expansions

Question

Lt (x->0) xcosx – log(1+x) x 2 how to solve these kind of questions without using expansions

Vikas TU · Answer

As it is 0/0 indeterminant form solve it by applying L-Hospital 2 times. Hence, => (cosx – xsinx – 1/(1+x))/2x => (-sinx – sinx -xcosx + 1/(1+x)^2)/2 Put x = 0 => ½.

seetaram dantu · Answer

what if it is not in the form of 0/0? like cosx- log(1+x)/x 2 ? if there is an other way too, please let me know

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Lt (x->0) xcosx – log(1+x) x 2 how to solve these kind of questions without using expansions

Lt (x->0) xcosx – log(1+x)
x²
^{how to solve these kind of questions without using expansions}

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Lt (x->0) xcosx – log(1+x) x 2 how to solve these kind of questions without using expansions

Lt (x->0) xcosx – log(1+x) x2how to solve these kind of questions without using expansions

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Lt (x->0) xcosx – log(1+x)
x²
^{how to solve these kind of questions without using expansions}