Differential Calculus> limits...

lim x→0 (e^tanx- e^x) / (tan x -x) is equal to?

(A) 1 (B) 1/2 (C) 1/3 (D) 0

[BOOK ANSWER: 1/2]

EXPLAIN .

Aditi Bhatnagar , 13 Years ago

Grade 12th Pass

3 Answers

Aman Bansal

Last Activity: 13 Years ago

Dear Aditi,

Expand tan x using taylors theorem and then reduce the expression to get the answer.

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Chetan Mandayam Nayakar

Last Activity: 13 Years ago

the Mcclaurin series expansion is

e^y=∑_{y=0 to ∞}yⁿ/n!

therefore, lim x→0 (e^tanx- e^x) / (tan x -x)=1+(1/(tanx-x))(tan²x-x²+tan³x-x³...

=1+(1/(tanx-x))(tan²x/(1-tanx))-(x²/(1-x))=1+tanx+x=1

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jitender lakhanpal

Last Activity: 13 Years ago

Dear Aditi

lim x→0 (e^tanx- e^x) / (tan x -x)

take e^xas common

so we get

lim x→0 e^x (e^tanx-x- 1) / (tan x -x)

this is nothing but the standard limit

limit x tends to 0 (e^x - 1)/x = 1

where x = tanx - x

so it is equal to 1

so the answer is 1 and not 1/2

we can also check this by L hospital rule as it is 0/0

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