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Numerade Educator



Problem 40 Medium Difficulty

Prove that $ \displaystyle \lim_{x \to 0^+}\sqrt{x} e^{\sin (\pi/x)} = 0 $.




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Video Transcript

this problem. Number forty of the Stuart Calculus eighth edition section two point three Proof that the Ltd's X approaches zero from the right of this court of X time E to the power sign of Pyrex is equal to zero. So with squeeze their own problems, we begin with a defect, usually concerning a Senate cosign function. In this case, we have a signed function. So we want to use that. And we know that sign as a function oscillators between pie or between negative one and positive one. So we write this fact down for us to use, we will use, or we will attempt to make the middle function look very similar to this function here almost exactly the same. And the next time it will take is to apply the exponential so exponentially ated to term, which gives us e to the negative one less than equal to either the sign of Pyrex. Uh, that's very cool to eat a positive one. And finally we will multiply by squared of X, which is appropriate because as we approach here on the right, this value always be positive. Nothing will change about this inequality. But as you will see. We will finally have the target function in the middle of this inequality, which is what we hope to have when using the squeeze there. Now, in order to no, with this limited's as experts zero from the right, we need to figure out what the limit is for the lower and the upper function as we approach x from the right, thanks as we approaching zero. All right, So for the upper function, we have screwed of X times e to the horn and six approaches here from the right, dysfunction approaches zero, and this is a constant. So this limit is equal to zero as X approaches zero from the right for the lower function squared of X. Either the negative one. Well, we have the same result because squared of X as a function of purchase zero. And this is a concept, right? So we've confirmed that the lower function and the upper function both approach zero as X approaches zero from the right. So by the squeeze serum, we can conclusively state that the limit is expert zero from the rain of the squared of X time's he to the power of sign. Pyrex is definitely equal. Two zero