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# 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.$\displaystyle \lim_{x\to 0^+} (\tan 2x)^x$

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So here the function that we're given is the limit. As X approaches zero from the right of the tangent of two X. Raised to the power of acts. We're going to want to rewrite this by taking the natural log of both sides. Um and the reason why we do this is so we can rewrite it now as the natural log of your tangent of two x Divided by one over X. Now with this, when we take the limit we get negative infinity over infinity. So we're gonna have to use low tells rule. Taking the derivative of the top and bottom. We end up getting a negative two X squared second squared two X over signed to Acts over Crossan Q. Acts that allows us to cancel some things um and simplify further. Um Ultimately though, when we apply the limit will get the indeterminate form again. So we have to take the derivative one more time. And that will give us negative for X. Over to co sign two X. Now, when we apply the limit is exposed zero, we get zero. So we see that the natural log of Y equals zero, which means that why is going to equal one for the final answer?

California Baptist University

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