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If it is given x=2, what is the value of (x^2 - 2^2 )/(x - 2). Do we get two answers here?
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We have the function f(x) = (x^2 - 2^2)/(x - 2). For x = 2, it is seen that f(x) takes the form (2^2 - 2^2)/(2 - 2) = 0/0 which is indeterminate.
To find the value of f(x) for x = 2, we have to use limits.
`lim_(x->2)((x^2 - 2^2)/(x - 2))` = `lim_(x->2)(((x-2)(x+2))/(x-2))` = `lim_(x->2)(x + 2)` = 4
The value of f(2) is not defined as such but as x tends to 2, f(2) tends to 4. Using limits, we can take the value of x as close to 2 as we want, and in this way it is determined that f(2) is equal to 4.
The value of (x^2 - 2^2)/(x - 2) for x = 2 is equal to 4.
Posted by justaguide on October 10, 2011 at 3:22 PM (Answer #1)
Salutatorian, Dean's List
Only one answer.its 4.
Posted by shihan on October 10, 2011 at 8:44 PM (Answer #2)
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