Limits are meant to be played with!

Calculus Level 3

lim ⁡ x → 0 e tan ⁡ x − e sin ⁡ x tan ⁡ x − sin ⁡ x = ? \large \lim_{x \rightarrow 0} \frac{e^{\tan x}-e^{\sin x}}{\tan x- \sin x} = ?

Find the limit above to 2 decimal places. If you think that the limit doesn't exist, enter 987.78 as answer.


The answer is 1.00.

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Arkajyoti Banerjee by Arkajyoti Banerjee , 20, India

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1 solution

lim ⁡ x → 0 e tan ⁡ x − e sin ⁡ x tan ⁡ x − sin ⁡ x \large \displaystyle \lim_{x \rightarrow 0} \frac{e^{\tan x}-e^{\sin x}}{\tan x- \sin x}

= lim ⁡ x → 0 e tan ⁡ x − 1 − e sin ⁡ x + 1 tan ⁡ x − sin ⁡ x = \large \displaystyle \lim_{x \rightarrow 0} \frac{e^{\tan x}-1-e^{\sin x}+1}{\tan x- \sin x}

= lim ⁡ x → 0 e tan ⁡ x − 1 − ( e sin ⁡ x − 1 ) tan ⁡ x − sin ⁡ x = \large \displaystyle \lim_{x \rightarrow 0} \frac{e^{\tan x}-1-(e^{\sin x}-1)}{\tan x- \sin x}

= lim ⁡ x → 0 tan ⁡ x e tan ⁡ x − 1 tan ⁡ x − sin ⁡ x e sin ⁡ x − 1 sin ⁡ x tan ⁡ x − sin ⁡ x = \large \displaystyle \lim_{x \rightarrow 0} \frac{\tan { x } \frac { { e }^{ \tan { x } }-1 }{ \tan { x } } -\sin { x } \frac { { e }^{ \sin { x } }-1 }{ \sin { x } } }{\tan x- \sin x}

= lim ⁡ x → 0 tan ⁡ x . lim ⁡ tan ⁡ x → 0 e tan ⁡ x − 1 tan ⁡ x − lim ⁡ x → 0 sin ⁡ x . lim ⁡ sin ⁡ x → 0 e sin ⁡ x − 1 sin ⁡ x lim ⁡ x → 0 ( tan ⁡ x − sin ⁡ x ) = \large \displaystyle \frac{\lim_{x \rightarrow 0} \tan { x } .\lim_{\tan x \rightarrow 0}\frac { { e }^{ \tan { x } }-1 }{ \tan { x } } -\lim_{x \rightarrow 0} \sin { x }. \lim_{\sin x \rightarrow 0}\frac { { e }^{ \sin { x } }-1 }{ \sin { x } } }{\lim_{x \rightarrow 0}(\tan x- \sin x)}

= lim ⁡ x → 0 tan ⁡ x . 1 − lim ⁡ x → 0 sin ⁡ x . 1 lim ⁡ x → 0 ( tan ⁡ x − sin ⁡ x ) = \large \displaystyle \frac{\lim_{x \rightarrow 0} \tan x.1 - \lim_{x \rightarrow 0} \sin x.1}{\lim_{x \rightarrow 0}(\tan x- \sin x)}

= lim ⁡ x → 0 ( tan ⁡ x − sin ⁡ x ) lim ⁡ x → 0 ( tan ⁡ x − sin ⁡ x ) = \large \displaystyle \frac{\lim_{x \rightarrow 0}(\tan x - \sin x)}{\lim_{x \rightarrow 0}(\tan x - \sin x)}

= lim ⁡ x → 0 tan ⁡ x − sin ⁡ x tan ⁡ x − sin ⁡ x = \large \displaystyle \lim_{x \rightarrow 0} \frac{\tan x - \sin x}{\tan x - \sin x}

= lim ⁡ x → 0 1 = 1 =\large \displaystyle \lim_{x \rightarrow 0} 1 = \boxed{1}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Can we Just take e sin ⁡ x e^{\sin{x}} common from the numerator ??

lim ⁡ x → 0 e sin ⁡ x ( e tan ⁡ x − sin ⁡ x − 1 tan ⁡ x − sin ⁡ x ) lim ⁡ x → 0 e sin ⁡ x ⋅ 1 = 1 \displaystyle\lim_{x \to 0} e^{\sin{x}} \left(\dfrac{e^{\tan{x}-\sin{x}}-1}{\tan{x}-\sin{x}}\right) \\ \displaystyle\lim_{x \to 0} e^{\sin{x}} \cdot 1 = \boxed{ 1 }

Sabhrant Sachan - 4 years, 11 months ago

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Nice one step solution, man. Hats off to you. I was desperate to solve this question anyway, that's why I had to go through so many steps.

Arkajyoti Banerjee - 4 years, 11 months ago

Could have used expansions.

Abhi Kumbale - 4 years, 9 months ago

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