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Computing Limits :In the earlier section we saw that there is a large class of function which allows us to use to calculate limits. However, there are also several limits for which it won't work easily. The reason of this section is to build up techniques for dealing with some limits that will not let us to just use this fact.
Let's firstly got back & take a look at one of the first limits which we looked at and calculate its exact value and verify our guess for the limit.
Example Evaluate the given limit.
Solution: Firstly let's notice that if we attempt to plug in x= 2 we get,
hence, we can't just plug in x = 2 to evaluate the limit .hence, we're going to have to do something else..
The first thing which we have to always do while evaluating limits is to simplify the function as much as possible. In this case that means factoring the numerator and denominator both. Doing this gives,Hence, upon factoring we saw that we could cancel an x - 2 from the numerator and the denominator both. Upon doing this now we have a new rational expression which we can plug into since we lost the division by zero problem. Thus, the limit is,
Note that it is in fact what we guessed the limit to be.
a) Let n = (abc) 7 . Prove that n ≡ a + b + c (mod 6). b) Use congruences to show that 4|3 2n - 1 for all integers n ≥ 0.
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