Question

Limits

Answer 1

\[\LARGE \lim_{x \rightarrow 0} \frac{e^{tanx}-e^x}{tanx-x}\]

Answer 2

LaTeX suggestion, use \tan rather than "tan"


`\tan ` is better than `tan` in LaTeX

\(\tan \) , \(tan\)

\(\LARGE \lim_{x \rightarrow 0} \frac{e^{\tan x}-e^x}{\tan x-x} \)

Answer 3

Okay

Answer 4

I think it is 1.

Answer 5

0/0 form appy lhospital rule

Answer 6

L hopsital didn't work for me

Answer 7

and @mathslover it is one

Answer 8

simply series expand e^x&e^tanx

Answer 9

\[\LARGE \frac{e^{tanx} \times \sec^2x -e^x}{\sec^2x-1}\]
after differentiating

Answer 10

still 0/0

Answer 11

And how can i expand using series for e^tanx? it would lead me nowhere,i tried that though.

Answer 12

@hartnn

Answer 13

e^x=1+x+x^2/2!+.......
e^tanx=1+tanx+(tanx)^2/2!+.....

Answer 14

yes

Answer 15

i don't think it will lead to the answer

Answer 16

Differentiate both numerator and denominator.

Answer 17

I did

Answer 18

yeah

Answer 19

\[\LARGE \frac{e^{tanx} \times \sec^2x -e^x}{\sec^2x-1}\]

Answer 20

we have to convert it in something like (e^x-1)/x.. this is equal to 1..

Answer 21

oops sorry, didn't scroll above. You're on right track, what you did next?

Answer 22

various method to solve limit problems

Answer 23

divide numerator and denominator by xe^x

Answer 24

but LHOSPITAL EASY METHOD..............

Answer 25

those were my first thoughts, donno whether it will work

Answer 26

L hospital is giving 0/0

Answer 27

@hartnn might work

Answer 28

Yes^

Answer 29

CONVERT DETEMINANT FORM

Answer 30

differentiate again..

Answer 31

DIFF AGAIN & AGAIN

Answer 32

lol

Answer 33

differentiate infinite times :P

Answer 34

there should be an easy alternate method for this

Answer 35

www.mathportal.org/calculators/calculus/limit-calculator.php?val1=(e^tan+x+-+e^x)+/+(tan+x+-x+)&val2=%5Clim%5Climits_%7Bx+%5Cto+0%7D+%7E+%5Cdfrac%7B%7B%5Cmathrm%7Be%7D%7D^%7B%5Ctan%5Cleft(x%5Cright)%7D-%7B%5Cmathrm%7Be%7D%7D^%7Bx%7D%7D%7B%5Ctan%5Cleft(x%5Cright)-x%7D&val3=0&rb1=both

even wolframs says the same but does NOT describe the answer.

Answer 36

Hmm, I think I got it. Wait

Answer 37

i told already it is one

Answer 38

Here it is.

\( \large {\lim _ {x \rightarrow 0 } \cfrac{e^{\tan x} - e^x}{\tan x - x} \\
\textbf{By taking e^x common from numerator and then using : } \\
\
\lim _ {x \rightarrow 0 } {f(x). g(x)} = \lim _ {x \rightarrow 0} * \lim _{ x\rightarrow 0} g(x) \\
\implies 1 ( \lim_ {x\rightarrow 0}\cfrac{e^{\tan x - x} -1}{\tan x -x} )\\
\textbf{Put tan x - x = y} : \\
\qquad \\
\lim _ {y\rightarrow 0} \cfrac{e^y - 1}{y} = 1 }\)

Answer 39

Sorry it took time for me to write it all. I am little bit tired now a days :)

Answer 40

Hope it helps @DLS

Answer 41

just one note,
as x-> 0
tan x-x ->0
and hence y->0
then you can apply that standard limit.

Answer 42

e^tanx-e^x/tanx-x= e^tanx[ 1- e^x/e^tanx]/ tanx-x
= e^tanx[ 1- e^x-tanx]/tanx-x
= e^ tanx[ 1- e^ -(tanx-x)]/tanx-x
further apply rule

Answer 43

DLS UNDERSTAND U

Answer 44

FINALLY ANS 1