Question

limits question........

Answer 1

OMG Kpop Fan?

Answer 2

\[\lim_{x \rightarrow 0} (e ^{^{tanx}} - e ^{x})/tanx - x\]

Answer 3

@pooja195 @Preetha

Answer 4

@Ashleyisakitty

Answer 5

@jim_thompson5910

Answer 6

@lionheart

Answer 7

\[is~your ~question~like ~this?\lim_{x \rightarrow 0}\frac{ e ^{\tan x}-e^x }{ \tan x-x }\]

Answer 8

yupp...

Answer 9

You're going to need to use l'hopitals :)

Answer 10

tan (0) = 0 which means this is indeterminate

Answer 11

then u need to make it determinant form...........the given answer is 1 but i am unable to solve it

Answer 12

\[\lim_{x \rightarrow 0}\frac{ e ^{\tan x}-1+1-e^x }{ \tan x-x }\]
\[\lim_{x \rightarrow 0}\frac{ \frac{ e ^{\tan x}-1 }{ \tan x }\times \tan x }{ \tan x-x }-\lim_{x \rightarrow 0}\frac{ e^x-1 }{ \tan x-x }\]
\[=\lim_{x \rightarrow 0}\frac{ \frac{ e ^{\tan x}-1 }{ \tan x } \times \frac{ \tan x }{ x } }{ \frac{ \tan x }{ x } -\frac{ x }{ x }}-\lim_{x \rightarrow 0}\frac{ \frac{ e^x-1 }{ x } } { \frac{ \tan x }{ x } -1}\]

Answer 13

\[\lim_{x \rightarrow 0}\frac{ e ^{\tan x}-1 }{ \tan x }=1\]

Answer 14

\[\lim_{x \rightarrow 0}\frac{ \tan x }{ x }=1\]

Answer 15

\[\lim_{x \rightarrow 0}\frac{ e^x-1 }{ 1 }=1\]

Answer 16

now you can complete it.

Answer 17

thanks a lot..

Answer 18

Wolfram agrees

Answer 19

i think we have done wrong
we have to use L hospital,s rule in between
but sorry now i am going to bed.

Answer 20

ur answer is correct lol.........