cu2l1b-squeeze-sandwich-theorem-limits-finite-pinch-calculus-ab-bc-i-ii-ap-limite-trigonometrico

step by step consideration of the basics of limits are presented in this video http://mathgotserved.com or http://youtube.com/mathsgotserved

Límite Trigonométrico

SANDWICH THEOREM

Good day students in this clip we are going to be going over some examples on how to use the sandwich theorem to find limits so before we start lets go ahead and take a look at the formulas that that I am going to show ahhm I am not going to be going over the detailed derivation of them there is a video installment on this sandwich theorem playlist that goes over how to derive this formula that I am showing you ok this uhm installment mainly goes over how to use these formula alright so the first formula is that the limit as x approaches 0 of sinx/x is equal to 1. ok in using this foundational formula we can also extend it to more situations ok so formula 2 the limit as x approaches 0 of sin ax/ax can be reduced to a over a which equals 1 ok so you can see how formula1 is a special case of formula 2 in the case of formula 1 a is equal to 1 alright ok number 3 the limit as x approaches 0 this is another formula you should master of sin ax/bx is equal to a over b just like this in this case the a these two are thesame, these two are different so you cant reduce it of sin ax over sin bx is equal to a over b similar to that one right there ok alright now there are two other limit ahhm limit ahhm sandwich theorem formulas that you need to know so number 5 the limit as x approaches 0 ahhm of x to the m sin 1/x to the n is equal to zero ok and number 6 its similar to that limit as x approaches zero of x to the n cosine one over x to the n this one is also equals to zero ok so note for five and six ahhm n and m have to be greater than zero ok alright so there you have it so these are the formulas that you should memorize in order to be able to do determine limits involiving the sand wich theorem quickly without any complications ok alright so lets apply these ahh formulas to some examples so the task is to determine the following limits determine the following limits alright problem number one what of if we want to find the limit as x approaches 0 of sin 5x/x ok you see how this form is consistent with formula number 3 in this case a is 5 and b is one so the answer is simply going to be a over b or 5 divided by 1 which is just five. alright there you have it lets take a look at question number two what is the limit as x approaches zero of ahm four x over sin 2x now this looks very similar to this case but it is kind of reciprocated alright so how do we do this? well we just reciprocate this then well take the limit, well reciprocate the value of the limit that we will get after evaluating what is on the inside ok so there are two ways we can do this I am going to use the reciprocal approach for this problem so what I am going to do is write this limit as the limit as x approaches zero of sin 2x over 4x so what did I just do I reicprocated this expression so let us raise that to the negative 1 ok so now we are going to take the limit of this expression inside the reciprocal symbol so this is going to become the limit as x approaches zero of 4x of sorry of uhm of sin 2x over 4x raised to the negative one power okay so what is this limit in here this limit is just like a over b sitiuation ax over bx its going to become two over four raised to the negative 1 two over four reduces to one half and the reciprocal of one half is two so there goes your limit okay. So this is one way of doing it just reciprocating this expression and then taking the limit of that and then reciprocating the result another way you can do this is you multiply the numerator and the denominator by 1 over 4x the numerator becomes one and this denominator becomes sin 2x over 4x and then you then have 1 over 2 over four and then you simplify and you have 1 over one half reciprocate and your answer will become 2 okay so that is another way of doing it. alright let us go to question 3 so how about if we want to find the limit as uhm x approaches zero of x to the third cosine one over x squared. what if we want to find this limit? Write this again one over x squared alright so if you take a look at this form it is consistent with formula number six this one right here so what do you think this limit is its simply zero. okay. so let me show you why this is the case. ok uhm using the sandwich theorem you do not have to do this i just want to show you why this is zero ok alright so uhm we are going to call this the explanation of number three. Explanation of three this is just fyi if you are curious this will help satisfy eh your curiosity explanation of number 3 so if take a look at number 3 you want to focus on the cosine expression we know that uhm cosine has a range of 1 and -1 the maximum output of cosine is 1 and the minimum is -1 so cosine is bounded between

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