This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
I have got another :), write it as:- x → 0 lim ( x ( e x − 1 ) ) 2 × e x 1 = 1 (Using lim x → 0 x e x − 1 = 1 ).
Approach 4: Use sinh ( x ) = 2 e x − e − x .
Log in to reply
Another approach: Note that the original form allows us to use L'Hopital's Rule. Thus, x → 0 lim x 2 e x + e − x − 2 = x → 0 lim d x d x 2 d x d ( e x + e − x − 2 ) = x → 0 lim 2 x e x − e − x = x → 0 lim 2 e x + e − x = 1
Log in to reply
It is incorrect to write d x d expression . You should, instead, write d x d ( expression ) . Also, d x d ( e x + e − x − 2 ) = e x − e − x = e x − e − x − 2 .
This is equivalent to Guillermo Templado's third method.
To the person who posted the question: please change the current n → 0 to x → 0 .
Log in to reply
Thanks , I have edited it. (Its surprising I didn't notice it O.o)
x → 0 lim x 2 e x + e − x − 2 = 1 .
Problem Loading...
Note Loading...
Set Loading...
1.- x → 0 lim x 2 e x + e − x − 2 = x → 0 lim x 2 1 + x + 2 ! x 2 + 1 − x + 2 ! ( − x ) 2 − 2 = 1 I have used e x ~ 1 + x + 2 ! x 2 when x → 0 applying McLaurin series
2.- x → 0 lim x 2 e x + e − x − 2 = x → 0 lim x 2 2 cos i x − 2 = x → 0 lim x 2 2 ( cos i x − 1 ) = x → 0 lim x 2 2 ( 2 ! − ( i x ) 2 ) = 1 I have used cos x ~ 1 − 2 ! x 2 when x → 0
3.- Use L'Hopital's Rule twice