solve {eq}lim _{x \to 0} \frac{(1+sinx)^{1/3} - (1 - 2 tanx)^{1/4}} { (sin x+tan^2x)} {/eq}

## Question:

solve {eq}lim _{x \to 0} \frac{(1+sinx)^{1/3} - (1 - 2 tanx)^{1/4}} { (sin x+tan^2x)} {/eq}

## L'Hospital Rule and its Idea:

If we want to find the limit {eq}\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} {/eq}

Which is of the form either {eq}\left( {\frac{0}{0}} \right) {/eq} or {eq}\left( {\frac{\infty }{\infty }} \right) {/eq}, then by using L'Hospital rule we can say that:

{eq}\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to a} \frac{{f'\left( x \right)}}{{g'\left( x \right)}} {/eq}

## Answer and Explanation: 1

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View this answerHere, in this case, the given limit can be calculated as:

{eq}\eqalign{ L& = li{m_{x \to 0}}\left[ {\frac{{{{(1 + sinx)}^{1/3}} - {{(1 -...

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Chapter 6 / Lesson 5Learn to define what the limit of a function is. Learn to describe the properties of limits and discover how to prove the properties of limits and see examples.