Limit at a Removable Discontinuity

Question

Let $$f$$ be the function defined by $$f(x) = \frac{(x+3)(x+2)}{(x+3)(x-3)}$$. The table below gives values of $$f$$ for selected values of $$x$$. What is the value of $$\lim_{x 	o -3} f(x)$$?

| x    | f(x)   |
|------|--------|
| -3.1 | 0.18   |
| -3.01| 0.168  |
| -2.99| 0.167  |
| -2.9 | 0.152  |

Accepted Answer

$$\frac{1}{6}$$

Answer

$$-\frac{1}{6}$$

Answer

When substituting x = -3, canceling the factor $$x+3$$ yields $$f(x)=\frac{x+2}{x-3}$$; substituting x = -3 gives $$\frac{-1}{-6}$$ which equals $$\frac{1}{6}$$.

Answer

$$0$$

Rank	User	Correct Count	Attempt Count	Total Time	Score
#1	meromaroseif	1	1	0m 00s	100
#2	humoyunmirzo027a	2	2	2m 40s	40
#3	kskim0724	0	1	0m 04s	-14
#4	lesleiman	0	1	0m 26s	-36

Limit at a Removable Discontinuity

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