<p>Okay I can't figure these out:</p>
<p>Why is the limit as x approaches 0 of [(2 - sq.rt. of 4 + x) / x] equal to -1/4?</p>
<p>Why is the limit as x approaches infinity of x * sin(1/x) equal to 1?? ( i thought it was supposed to be zero)</p>
<p>Aghhh please help me :(</p>
<p>You use L’hospital’s rule. For number one you don’t need to rewrite it before applying L’hospital. After L’hospital’s rule, you have -.5(4+x)^-.5, which is -1/4 after you substitute x=0.</p>
<p>For number two you have to rewrite it so it’s in an indeterminate form; I suggest writing it as sin(1/x)/(1/x). After L’hospital, you have -cos(1/x)(1/x^2)/(-1/x^2), and after substituting x=inf you get 1.</p>
<p>^Do realize that L’Hopital’s Rule is not part of the Calc AB curriculum.</p>
<p>Oh, then I have no idea how to solve these limits without using it. The only other way I can think of is graph it and see what the limit is graphically.</p>
<p>For number 1, you can multiply the top and bottom by the conjugate: 2 + sq.rt. of 4 + x</p>
<p>…I’m not sure how to solve the other one, sorry.</p>
<p>@314159265: L’Hopital Rule is included in the AP Calculus AB curriculum. I’ve seen some problems on the AB test that requires L’Hopital’s Rule to solve it. </p>
<p>Its a major lifesaver for solving limits if you have knowledge about it.</p>
<p>^Really? Oops. </p>
<p>Still, regardless of whether or not it’s in the curriculum, L’Hopital’s Rule is amazing (and easy too). Just learn it!</p>
<p>Actually, 314159265 is right. L’Hopital’s Rule is not required for any limits on the AP Calculus AB exam. It is an AP Calculus BC topic exclusively.</p>
<p>For the limit as x -> infinity of x<em>sin(1/x), you have to recall that the limit as x approaches 0 of sin(x) / x = 1. (This is a limit that many instructors have students memorize early in the year, as you use it later on to prove that the derivative of sin x = cos x.) The trick is to re-write x as 1/(1/x). Then the lim x-> infinity of x</em>sin(1/x) = lim x->infinity of sin(1/x) / (1/x). Let u = 1/x. Then as x->infinity, u->0, so lim x->infinity of sin(1/x) / (1/x) = lim u-> 0 of sin(u)/u, which we established earlier as 1.</p>
<p>^I’m not sure if that’s the official AB solution since I don’t think many people memorize that limit (and since the value of that limit is derived from L’Hospital). I never even saw that limit until we reached L’Hospital, and it was just a problem out of the problem set; nothing to memorize.</p>
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<p>Its not exclusive to AP Calculus BC; its also found (and applied) on AP Calculus AB too, even though it may not be listed explictly on the AP Calculus AB curriculum found on the College Board’s website. Though, I have to say, learning it in AB does save your butt in certain situations; its a very nice rule to use and abuse. </p>
<p>The one that’s found in BC covers more in depth than AB (and previous knowledge of the basic L’Hopital Rule is a good thing to have before learning more about it).</p>
<p>My AP Calculus AB class covered L’Hopital Rule. Takes one day to learn, but it can be the key to unlocking certain limit problems that seems unsolvable.</p>
<p>^ You’re allowed to use it on the AP Calculus AB exam, but no AB exam question requires its usage.</p>
<p>^^ Actually, the value of that limit is derived from the Squeeze Theorem, if you really want to find it algebraically.</p>
<p>The proof that d/dx[sin x] = cos x explicitly relies on you knowing this particular facet. (See [Derivative</a> of the Sine Function](<a href=“Oregon State University”>Oregon State University) for an example of this.) The problem is, if you use L’Hopital’s Rule to prove that lim x -> 0 sin x / x = 1, then you need to know the derivative of sin x to find that. In order to find the derivative of sin x, you need to know the limit. So one of those has to be able to come first. :)</p>
<p>And to be fair, the AP Exam doesn’t ask questions that require you to have that limit memorized on the AB exam.</p>
<p>We did memorize that the limit as x approaches 0 of sin (x)/ x is 1 (but not l’hopital’s rule), but I wasn’t sure exactly how to use that to solve the problem. I had tried multiplying the conjugate for the other one, but it didn’t work for me the first time before I posted this. WOW I’m so good at calculus</p>
<p>Sorry for taking so long to reply guys, but I really do appreciate the help. Got a 91 on the test!</p>