Limits removable discontinuities 1.3 ap calculus ab bc ib calc radical trig squeeze theorem wksht

sine five X is sine square 5X in the X times X is X square in the bottom here right so now all I like to go ahead and down to make this I can probably these two limits with this formula right here we have on a the limit as X approaches zero of sine X over BX is a repeat so we have A1 here anyone here so let's go ahead and distribute this limited is factors is one of the properties of limits this problem now becomes one over for times the limit as X approaches zero of sine 5X over one over one X times the limits is X approaches 0+ 5X over one okay so this limit simply becomes 501 in this one also becomes five over one so our limits is going to become one over four times one times five over one when you multiply across you final answer will be 25 over four okay so that's that right let's take a look at question two is simple radicals so have to rationalize here if we just evaluate this limit by direct substitution will get any indeterminate form because will have you zero in the denominator okay so on the what we going to do here is rationalize the numerator so multiply the numerator by the square root of X -2+ one over the square root of X -2+ one so this is the conjugate okay so the distribute in the numerator will have the limit as X approaches 3X -2 minus one divided by the denominator will have X -3 times the square root of X -2+ one in the numerator all I did is I took advantage of the difference of squares formula you have a's in a minus B times a plus be that's can become a square minus T-square okay so when I squared is square roots of X -2 I have X -2 when I square one one there is go ahead and simplify further will have the limit as X approaches three of X -3 divided by the quantity X -3 times the square roots of X -2+ one of these two are identical so can divide the numerator the denominator by X -3 so we now have the limit as X approaches three of one over the square roots of X -2+ one now we can on about with this limits by direct substitution plug-in three will have one over the square roots of 3-2+ one way it becomes one over 3-2 is one so the squared of one is one that becomes one over one plus one which he calls to one over to is right let's take a look at question number three we are to find the limit as X approaches zero of one minus cosine 2X divided by cosine 4X minus one before we about we just limit first so let's go over identity that Ali using here which is the double angle identity for cosine okay so cosine to outline is equal to two cosine square out the minus one right we can apply that situation to the denominator here we have to cosine to times two out okay apply the formula above here this becomes two cosine square to all the minus one the same daily differences that the double angle is doubled here and then that is have the double angle directly okay right now let's take a look at this limit as indicated earlier you can apply the double angle identities for cosine to this term here cosine 4X let me write it in a way that makes it obvious so you have the limit as X approaches zero of one minus cosine 2X now that divided by divided by cosine to times 2X -15 so let's apply the double angle identity to the denominator think about our plus 2X here so this can be out the all limit now becomes the limit as X approaches zero of one minus cosine 2X divided by no in the denominator you going to have to cosine square 2X minus one okay so here and down so can be to cosine square 2X imagine is still 2X is your also okay so you have to cosine square off a minus one but out what these 2X so that's why we have it here right so is minus one indelibly -1 there before so this issue is that into this that