How to read limits?

A limit means that the function is APPROACHING that value. It may not actually reach that value; it may only get close to it.
 


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Your answers on exercises 1, 6, and 7 are correct.

Your answers on exercises 2, 3, 4, and 5 are not correct.

On exercise 2, look at the graph of f(x). As you move from RIGHT to LEFT, approaching x = 1 from the RIGHT, what value is f approaching?

On exercise 3, look at the graph of f(x). As you move from LEFT to RIGHT, approaching x = 0 from the LEFT, what value is f approaching?

On exercises 4 and 5, your result of 0/0 is not a number. Whenever a limit exists, it must be a Real number. If you end up with something over zero, then the question of whether or not the limit exists is not yet answered. We refer to this result as "indeterminant".

Therefore, you must find another way to evaluate the limit or show that it does not exist. Usually, the goal is to find a way to rewrite the given expression so that direct substitution does not result in a denominator of zero.

On exercise 4, factor the given denominator (it's a difference of squares). You will see a cancellation. After you simplify by cancelling, you may use direct substitution.

On exercise 5, multiply top and bottom by the conjugate of the denominator (i.e., rationalize the denominator). Then use direct substitution.

 
Thank You, mmm4444bot . I think I got it.


2) Moving from Right to Left, approaching x = 1 . The answer is 2? which is not in the answer key so it's , "None of the above"

3) Moving from Left to Right, approaching x = 0. The answer is 1.

4) Factoring out the denominator of (x^2 - 4) I get ((x - 2) / (x+2)(x-2). The X-2 cancels out.. Leave me with 1/(x+2) substituting x with 2. I get 1/4..

5) Using the conjugate of the denominator.
(sqrt(x) +2 ) /1 ---> (sqrt(4)+2) /1 ---> (2+2 / 1) ---> 4/1 ---> 4


Am I correct? Thank you, for helping me out, mmmm4444bot
 
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