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evaluate the limit. $$ \lim _{x \rightarrow 0} …

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Problem 28 Medium Difficulty

evaluate the limit.
$$
\lim _{x \rightarrow 0}\left(\cot x-\frac{1}{x}\right)
$$

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Video by Bahar Tehranipoor

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Calculus 1 / AB

Calculus for AP

Chapter 4

APPLICATIONS OF THE DERIVATIVE

Section 5

L'Hopital's Rule

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Video Transcript

For this question, we're going to evaluate the limit and what we can do is use hopital to combine this into x, co, tangent, x, minus 1 over x, and if we take our limit expression 0 we're going to get infinity over infinity, which is indeterminate. We can use a petal, so the derivative of f, which is going to be numerator g being a denominator, is going to be x, negative, x, co, secant squared x, plus co tangent x, while the derivative of g is just going to be 1, so we can Rewrite our limit as a limit of x, approaching 0 of co, tangent x minus x. Now what we can do. We can also rewrite this with sin denominator. So we'll do co. Tangent of x, minus x, cosecant square of x can be rewritten as co sine x over sine x minus x over sine squared x. So if we make this sine squared x, we multiply sine here. We write this as sine x, cosine x, minus x, over sine squared x. Now, taking the limit of this, we still can't do it we're going to get easier in the denominator, we'll differentiate again so in the bottom, we're going to get the derivative of sine squared x is just going to be 2 sine x, cosine x, derivative of sine X, cosine x, we can actually do a trick with trig and place 1 half sine of 2 x. Take that thrives going to give us a cosine of 2 x derivative of x is 1 to minus 1 point. So the limit of this will still give us a determinate form so 1 more thing we to do. We can make this sine of 2 x and differentiate again, so derivative cosine 2 x is just going to be negative. 2 sine of 2 x derivative of sine of 2 x is going to give us 2 cosine 2 plugging in our x approaching 0. Now we're going to see successfully we're going to get 0 numerator but 1 in the denominator. So our limit evaluates out to 0.

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