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Find the limit or show that it does not exist.

$ \displaystyle \lim_{x \to \infty} (e^{-2x}\cos x) $

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Okay we want to find the limit of this function as X approaches infinity. One of the first things I'd like to do is I want to rewrite E to the negative two X. As one over E to positive two X. Uh So each of the negative two X. Does equal one over E. To positive two X. So we want to find a limit of one over E. To the two X. Times Cosine of X as X approaches infinity. Now the absolute value of one over E. To the two X times Cosine X is less than or equal to the absolute value of one over E. To the two X times the absolute value. Um Co Synnex. So the absolute value of a product is less than or equal to the product of the absolute values. Okay so the absolute value of one over E 22 X times. Cosine X is going to be less than or equal to the absolute value of one over E 22 X times the absolute value of co sign X. Now if you remember your co sign function when you grab the Cosine function uh it oscillates between one and negative 11 and negative one. Uh So cosine X. Because it's always bounded by one and negative one. The absolute value of co sign of X will always be less than or equal to one. Um If co sign of X is negative one the absolute value will be one. Uh coastline effects is one than the actual value of coastline effects is one. Other than that Co sign of X is in between one and negative one for example, one half. Uh and the absolute value of one half is one half. Long story short, the absolute value of cosine of X will always be less than or equal to one. So uh the absolute value of one over E to the two X times the absolute value of cosign fx is going to be less than or equal to the absolute value of one over E to the two X times why? Because since the absolute value co sign of X is less than or equal to one than the absolute value of one over E to the two X times the absolute value cosign X has to be less than or equal to the absolute value of one over E to the two X times one. Because the absolute value of co sign of X uh is always less than or equal to one. So you're always going to be times in Buy something that's less than or equal to one. No, let's look at this expression as X goes towards infinity. As X goes towards infinity, two x goes towards infinity and E Raised to the two x goes towards infinity. So this entire uh, denominator, okay, as X is going towards infinity, two times X goes towards infinity. And so E raised to a very large number. Uh as X goes towards infinity, two times X goes towards infinity. E raised to the two times X. Uh this will all go towards infinity. So the denominator uh here will go towards infinity as X goes towards infinity. Well, if the denominator is getting increasingly larger and larger with outbound, if the denominator is approaching infinity than one over E to the two X one over this, denominator must be approaching zero. As the denominator gets larger and larger and larger, one over the denominator gets smaller and smaller. For example, one over 1000 is small, one over a million is even smaller, one over a billion Is even smaller than that. You're getting closer to zero as X is approaching infinity. And so this entire uh fraction the one over E to the two X is going to approach zero. And so the absolute value of one over E. To the two X times Cosine of X was less than or equal to this amount here, the one over E 22 X times one. The absolute value of one over each of the two x times Cosine of X is less than or equal to the absolute value of one over E 22 X times one. Well, this fraction the one over E 22 X uh will approach zero as X approaches infinity and as this amount approaches zero, the absolute value of that amount will still be approaching zero and a number that approaches zero times one will still be a number approaching zero. So when x is very large, uh the absolute value of one over each of the two X times co sign X. Uh this is going to approach zero. And so if the absolute value of one over each of the two X times cosine of X is approaching zero, that means maybe this is positive, maybe it's negative for different values of X. Uh But as X is approaching infinity, The absolute value, the magnitude of this amount approaches zero. And so at the magnitude of this amount approaches zero, The absolute value of this amount approaches zero. Then the one over E to two X times cosign effects must be approaching zero and so one over E to the two X times Cosine of X. This entire expression approaches zero as X approaches infinity. So the limit of each and negative two X times Cosine of X. As X approaches infinity is zero.