Evaluate the Limit

Calculus Level 3

Try without using L'HOSPITAL Rule

-e/4 3e/2 -e/2 e/2

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1 solution

Ronak Agarwal
Jul 8, 2014

O u r r e q u i r e d l i m i t i s l i m x 0 e e l n ( 1 + x ) x t a n ( x ) = e ( l i m x 0 ( 1 e l n ( 1 + x ) x x ) x l i m x 0 t a n ( x ) x ) A p p l y i n g s e r i e s e x p a n s i o n w e g e t l n ( 1 + x ) x x x / 2. H e n c e h e h a v e ( e / 2 ) ( l i m x 0 ( 1 e x / 2 ) x / 2 l i m x 0 t a n ( x ) x ) = e / 2 Our\quad required\quad limit\quad is\quad \underset { x\rightarrow 0 }{ lim } \quad \frac { e-{ e }^{ \frac { ln(1+x) }{ x } } }{ tan(x) } =e(\frac { \underset { x\rightarrow 0 }{ lim } \frac { (1-{ e }^{ \frac { ln(1+x)-x }{ x } }) }{ x } \quad }{ \underset { x\rightarrow 0 }{ lim } \quad \frac { tan(x) }{ x } } )\\ Applying\quad series\quad expansion\quad we\quad get\quad \frac { ln(1+x)-x }{ x } \approx x/2.\\ Hence\quad he\quad have\quad (e/2)(\frac { \underset { x\rightarrow 0 }{ lim } \frac { (1-{ e }^{ x/2 }) }{ x/2 } \quad }{ \underset { x\rightarrow 0 }{ lim } \quad \frac { tan(x) }{ x } } )=e/2

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