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Find the limit. $ \displaystyle \lim_{x \to 0} \…

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Problem 43 Medium Difficulty

Find the limit.
$ \displaystyle \lim_{x \to 0} \frac {\sin 3x}{5x^3 - 4x} $

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Video Transcript

All right, our goal is to solve this limit problem and notice that, if i plug in 0 to the top, we can look both the numerator and denominator separately, just to see what would happen so if we plug in basically, if we do, the limit as x Goes to 0 for the numerator we get sine of 0, which is just 0, so top looks good. We can find that limit. Now, let's see what happens if we look at just the bottom, if we look at just the bottom, then we will have 5 x cubed minus 4 x as our denominator. If we take the limit, as x, goes to 0, we can try plugging in 0 and we get 0 as well. So now we have what is known as a 0 over 0 case, meaning i can apply. Lopetal'S rule and tobit's rule is really amazing. It basically says that if we have the conditions met, as you already see, then in this case, if i have, the limit as x goes to 0, for example, of f of x over g of x. So long as both are individually 0, as we already proved, then this will equal. The limit as x goes to 0 of derivative over derivative and notice. This is not potient rule. This is simply lupito's rule derivative over derivative and then we can do the limit. So this is known as lopital's rule and some books say the hospital's rule, but french is lupito's, opal's rule all right. So, let's just kind of mark that off now to do lupito's rule you have to have individually the limits top and bottom are both in this case 0. So it works okay. So that means we can keep going and we can say: okay, we do have that 0 over 0 case. So in the limit as x goes to 0, then it should be the same as a derivative of the top over the derivative of the bottom. Well. Derivative of the top is cosine of 3 x, but we have chain role, so we still have to multiply by 3. The bottom we can do power will just take its derivative, so we get 15 x squared minus 4 point. Now, let's go ahead and see what happens if we plug in 0, if we plug in 0 to the top, we get cosine of 0, which is 1 times 3, so we just get 3 and on the bottom, we'll get 0 minus 4. So minus 4 point. So we have solved this limit and the answer to this limit is minus 3 quarters by use of lobito rule are in, he help have a wonderful day.

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Calculus: Early Transcendentals

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