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Find $\lim _{x \rightarrow 0} \frac{x-\sin x}{x^{3}}$
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Chapter 1
Practice Test 1
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So for this problem, we know that we want to use low petals rule. That means that we're just going to take the limit as X approaches zero of the derivative of the top and bottom independently. I'm so ddx of our our numerator. Um, it's just gonna be the derivative of X, which is one minus the derivative of fine X will be minus coz I knew X over the derivative with respect to X of the denominator is going to be over three x squared. And so from here using Loki told rule, we would just plug in X is equal to zero in salt this way. So that would be one minus the coast. I'm of zero over three times zero and so we can see that this is still going to be an indeterminant form because it's going to be zero over zero. So that means that we're going to need to use Loki tolerable again. I'm so essentially we're just going to take the derivative of the top and bottom independently. Once again, I'm so the derivative of one is just a constant and the derivative of negative coastline X is just going to be sine X over the derivative of three X Squared is six x, but we can see that once again. If we play a game. X is equal to zero. This is just going to be zero over zero. So again it's indeterminate form, so we're going to have to use, Libby told Rule one more time to get rid of this six x, I'm so again taking the derivative, we're going to have a limit as H approaches zero of the derivative of Sine X, which is co signed X over the derivative of six Ex, which is just six. And finally we're going to be able to plug in H is equal to zero. Here, Amore X is equal to zero. I'm not sure why, but h there. But so that's going to be equal to the coast side of zero over six. So you know, the coastline of zero is one. That means that our limit as X approaches zero
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