Evaluate a) ( lim x ightarrow 1 frac1-cos (5 x)3 x ) b) (...

Evaluate a) ( lim _{x ightarrow 1} frac{1-cos (5 x)}{3 x} ) b) ( lim _{x ightarrow infty} frac{ln x}{x} ) c) ( lim _{x ightarrow 0}left(frac{1}{x}-frac{1}{sin x} ight) ) d) ( lim _{x ightarrow infty} x^{3} e^{-x} )

Transcript

00:01 Let's evaluate these limits.

00:02 The first one we have the limit as x goes to 1 of 1 minus cosine 5x over 3x and let's just plug in 1.

00:18 So 1 minus 5 times 1 is 5 over 3 and the point x equals 1 exists so this is just going to be our answer here right we just plug it in and this is what we get there's nothing special about this point okay so the next one is a little bit trickier it is the limit as x goes to infinity limit as x goes to infinity of ln x over x okay so there's a couple ways we could do this the first one is well we know that one over x i mean yeah x let's just talk with the denominator x goes to infinity like this steadily and relatively quickly whereas ln x looks a little more like this right so it goes towards infinity but very very slowly so the top goes to infinity slower than the bottom goes to infinity so right there you could kind to figure out that that's going to be equal to zero but if you still want to try to solve it properly anyway you could take the derivatives of the top and the bottom so the derivative of the top ln x is one over x and the derivative of the bottom is one so the limit as this goes to infinity is the same thing as the limit as one over x goes to infinity and at this point you you can see as the bottom gets infinitely large, this number approaches 0.

01:51 So the answer for the second one is going to be 0.

01:55 Okay, so for this one, we have the limit as x approaches 0, 1 over x minus 1 over sine x.

02:05 So the first thing i'm going to do here is to combine these two fractions.

02:11 So that is going to give us the limit as x approaches 0 of, let's see, this is going to be multiply this by sine x, multiply this by x over x sine x, right? and then when we plug in 0, this doesn't make any sense at all.

02:34 So i'm going to use l 'hopital's rule again, and i'm going to take the derivative of the top.

02:40 So each one of these is going to be evaluated separately so sine x is cosine x minus 1 so that's already a lot better now down here we have a product rule so we have uh the derivative of the first one times whoops the second one plus the derivative of the second one times the first one okay all right so we still get zero here because you know this this so i'm going to do the same thing again i'm going to take the derivative once again so this is going to be limit as x approaches zero derivative of the top that is going to give us negative sine x over okay now let's look at the bottom here we're going to have a few things going on sine is cosine x the second one is going to be product rule once again so one cosine x plus x x, this is going to be a negative sign, so this is going to be a minus.

03:48 Okay, and now we can plug in 0, right? because this becomes 0, but this and this are going to be 1.

03:55 So negative sine of 0 over cosine 0 plus cosine 0.

04:04 We get 0 over 2, which is the same thing as 0...

Question

Evaluate a) ( lim _{x ightarrow 1} frac{1-cos (5 x)}{3 x} ) b) ( lim _{x ightarrow infty} frac{ln x}{x} ) c) ( lim _{x ightarrow 0}left(frac{1}{x}-frac{1}{sin x} ight) ) d) ( lim _{x ightarrow infty} x^{3} e^{-x} )

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