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Numerade Educator

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Problem 65 Hard Difficulty

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^+} (4x + 1)^{\cot x} $

Answer

$$\begin{array}{l}
y=(4 x+1)^{\cot x} \Rightarrow \ln y=\cot x \ln (4 x+1), \text { so } \lim _{x \rightarrow 0^{+}} \ln y=\lim _{x \rightarrow 0^{+}} \frac{\ln (4 x+1)}{\tan x} \stackrel{\Perp}{=} \lim _{x \rightarrow 0^{+}} \frac{\sqrt{4 x+1}}{\sec ^{2} x}=4 \Rightarrow \\
\lim _{x \rightarrow 0^{+}}(4 x+1)^{\cot x}=\lim _{x \rightarrow 0^{+}} e^{\ln y}=e^{4}
\end{array}$$

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Video Transcript

For this problem. We're gonna be taking the limit as X approaches zero from the right four X plus one to the cook hand in events. However, we're going to simplify this by taking the natural log of both sides and getting it into a nicer form. That form is going to be the natural log of four X plus one provided by the tangent of X. Now, when we evaluate the limit as X approaches zero from the right, we get 0/0, which is um the indeterminant form. But this is a good thing though because it allows us to use locals rule taking the derivative of the numerator, we get 4/4 X plus one and then taking the derivative of the denominator. We get second squared X. Um and we Seacon square decks, we now can evaluate the limit. When we plug in zero with this, we now get um four because we can rewrite this. So we would get four as our answer. But remember that this is not just for its Y equals the natural log before, or the natural log of Y. E equals four. So when we solve for why we get E to the fourth, so I need to, the fourth is going to be our final answer.