How do I remove these discontinuities?

**iluvmathbutitshard** · September 24th 2010, 07:53 AM

So I understand how to find where a discontinuity is, but I am not sure how to remove it. Any tips?

Question 1: Identify and remove the discontinuities of f(x) = [sin(x+1) / tan(x+1)]
a) discontinuity at x = -1, remove by making f(-1)= 1
b) discontinuity at x = -1, can not removed
c) discontinuity at x = 0, remove by making f(0) = 1
d) discontinuity at x = 0, remove by making f(0) = 1

Question 2: Identify and remove the discontinuities of f(x) = [(x^3 - 3x^2 + 2x - 6) / (x - 3)]
a) discontinuity at x = 4, remove by making f(3) = 9
b) discontinuity at x = 4, can not remove
c) discontinuity at x = 3, remove by making f(3) = 11
d) discontinuity at x = 4, remove by making f(4) = 11

Any help is appreciated.

**Soroban** · September 24th 2010, 08:26 AM

Hello, iluvmathbutitshard!

So I understand how to find where a discontinuity is,
but I am not sure how to remove it. Any tips?

(1) Identify and remove the discontinuities of: . $f(x) \:=\:\dfrac{\sin(x+1)}{\tan(x+1)}$

(a) discontinuity at $x = \text{-}1$ , remove by making $f(\text{-}1)= 1$
(b) discontinuity at $x = \text{-}1$ , cannot removed
(c) discontinuity at $x = 0$ , remove by making $f(0) = 1$
(d) discontinuity at $x = 0$ , remove by making $f(0) = 1$

When $x = \text{-}1$ , we have: . $f(\text{-}1) \;=\;\dfrac{\sin0}{\tan0} \;=\;\dfrac{0}{0}$

The function is discontinuous at $x = \text{-}1$

Note: . $\dfrac{\sin(x+1)}{\tan(x+1)} \;=\;\dfrac{\sin(x+1)}{\frac{\sin(x+1)}{\cos(x+1)} } \;=\; \cos(x +1)\;\text{ provided }x \ne \text{-}1$

The graph is the curve: $y \:=\:\cos x$ with a "hole" at (-1, 1).

We find that: . $\displaystyle \lim_{x\to\text{-}1}\cos x \;=\;1$

If we let $f(\text{-}1) \,=\,1$ , we can "plug the hole".

The answer is (a).

(2) Identify and remove the discontinuities of: . $f(x) \:=\: \dfrac{x^3 - 3x^2 + 2x - 6}{x - 3}$

(a) discontinuity at $x = 4$ , remove by making $f(3) = 9$
(b) discontinuity at $x = 4$ , cannot removed
(c) discontinuity at $x = 3$ , remove by making $f(3) = 11$
(d) discontinuity at $x = 4$ , remove by making $f(4) = 11$

The function is: . $f(x) \;=\;\dfrac{(x-3)(x^2+2)}{x-3}$

When $x = 3$ , we have $f(3) \,=\,\dfrac{0}{0}$

The function is discontinuous at $x = 3.$

Note: . $f(x) \:=\:\dfrac{(x-3)(x^2+2)}{x-3} \:=\:x^2+2 \;\text{ provided } x \ne 3.$

The graph is the curve: $y \:=\:x^2+2$ with a "hole" at (3, 11).

We find that: . $\displaystyle \lim_{x\to3}(x^2+2) \:=\:11$

If we let $f(3) = 11$ , we can "plug the hole".

The answer is (c).

**HallsofIvy** · September 25th 2010, 04:18 AM

You should understand that discontinuities are only rarely "removable".

For f(x) to be continuous at a point, x= a, three things must be true:
1) f(a) must exist.
2) $\displaytype\lim_{x\to a} f(x)$ must exist.
3) $\displaytype\lim_{x\to a} f(x)$ must be equal to f(a).

If (2) is true, if $\displaytype\lim_{x\to a} f(x)$ exists, but (1) or (3) or both are not, we can make f continuous at x= a by defining $\displaytype f(a)= \lim_{x\to a} f(x)$ . If (2) is not true, there is no way to make f continous at x= a.

What Soroban did was find the limits at each point and define the function to have that value.

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