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leejiawen leejiawen
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10 years ago
f(x)=(x+18) (lx+12l / x+12), c=-12

the l l signs are suppose to be absolute value signs.
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jtrieu2jtrieu2
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10 years ago
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wrote...
10 years ago
Consider x < -12, the limit will be 6(-1) = -6 since (lx+12l / x+12) = -1 for x< -12.

Consider x > -12, the limit will be 6(1) = 6 since (lx+12l / x+12) = 1 for x > -12.

Since the lilmits from the two sides are not equal we say the limit does not exist at c = -12.  The graph of f(x) will have a jump discontinuity at x=-12.

Written piecewise f(x) = -x-18 for x<-12 and f(x) = x+18 for x>-12.
wrote...
10 years ago
Neither of the previous two answers really does it right, though they get the right answer. Taking limits is about considering variations that TEND TO ZERO.  In the present case, consider the case of a value of x given by:

x = -12 + D

where D is a small number. Then the function evaluates to:

f(x+D) = (6+D) x |D| / D

As a previous answerer implied, if D is positive, the function evaluates to 6+D. Then, AS WE CONSIDER SMALLER AND SMALLER VALUES OF D, the value tends to 6 as D tends to 0. But if D is negative, the function evaluates to -(6+D). Then, as we consider smaller and smaller magnitudes of D, the value tends to -6 as D tends to 0.
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