`lim_(x->0) (e^x - e^(-x) - 2x)/(x - sin(x))` Find the limit. Use l’Hospital’s Rule where appropriate. If there is a more elementary method, consider using it. If l’Hospital’s Rule...

`lim_(x->0) (e^x - e^(-x) - 2x)/(x - sin(x))` Find the limit. Use l’Hospital’s Rule where appropriate. If there is a more elementary method, consider using it. If l’Hospital’s Rule doesn’t apply, explain why.

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Chapter 4, 4.4 - Problem 38 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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nees101 | Student, Graduate | (Level 2) Adjunct Educator

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Given the limit function `lim_{x->0}(e^x-e^(-x)-2x)/(x-sin(x))` . We have to find the limits.

Applying the limits we get,

`lim_{x->0}(e^x-e^(-x)-2x)/(x-sin(x))=0/0`

So we have to apply the L'Hospital's rule and again applying the limits we get,

`lim_{x->0}(e^x+e^(-x)-2)/(1-cos(x))=0/0`

Applying l'Hospital's rule again we get,

`lim_{x->0}(e^x-e^(-x))/sin(x)=0/0`

Therefore we have to apply the L'Hospital's rule again, i.e.

`lim_{x->0}(e^x+e^(-x))/cos(x)=2`

Hence the limit is 2.

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