# Thread: Continuity (Missing constant)

1. ## Continuity (Missing constant)

*THis was a question on my midterm today, and I probably got it wrong and was confused... The question was something like this.

Given: (i was given 2 graphs for functions F and G)

1. The first graph is of function F. The limits as x approaches 1 from the right of f(x) is 3 while from the left of f(x) is 1. (No values are defined at x =1 )

2. Another graph is of function G. G is defined at x =1 and it's equal to limit as x approaches one from the left of g(x) is -2 (defined) while from the right it is 2 (not defined, a hole).

3. I was given that (f*g)(fg) = f(x)g(x) is not continuous because x=1 for function F is not defined

Question:

If h(x) = { f(x) if x is not equal to 1
'a' if x is equal to one } < Piece-wise of h(x)

Is there a value for 'a' that makes h(x)g(x) continuous? If not, explain why. If yes, explain and include limits to further support your reasoning.

My answer:

Well, my reasoning is that we know the f and g is not continuous so that makes h(x) which is composed of f(x) is not continous. This means that the proudct of h(x) and g(x) is not continuous, so there is no value of 'a' that makes them continuous. I'm pretty sure this is wrong, but i had no clue!

2. ## Re: Continuity (Missing constant)

Originally Posted by lc99
*THis was a question on my midterm today, and I probably got it wrong and was confused... The question was something like this.

Given: (i was given 2 graphs for functions F and G)

1. The first graph is of function F. The limits as x approaches 1 from the right of f(x) is 3 while from the left of f(x) is 1. (No values are defined at x =1 )

2. Another graph is of function G. G is defined at x =1 and it's equal to limit as x approaches one from the left of g(x) is -2 (defined) while from the right it is 2 (not defined, a hole).

3. I was given that (f*g)(fg) = f(x)g(x) is not continuous because x=1 for function F is not defined

Question:

If h(x) = { f(x) if x is not equal to 1
'a' if x is equal to one } < Piece-wise of h(x)

Is there a value for 'a' that makes h(x)g(x) continuous? If not, explain why. If yes, explain and include limits to further support your reasoning.

My answer:

Well, my reasoning is that we know the f and g is not continuous so that makes h(x) which is composed of f(x) is not continous. This means that the proudct of h(x) and g(x) is not continuous, so there is no value of 'a' that makes them continuous. I'm pretty sure this is wrong, but i had no clue!
$\displaystyle \lim_{x \to 1^-} \left[ h(x)g(x) \right] = \left(\lim_{x \to 1^-} h(x)\right) \left( \lim_{x \to 1^-} g(x) \right) = (-1)(2) = -2$
$\displaystyle \lim_{x \to 1^+} \left[ h(x)g(x) \right] = \left( \lim_{x \to 1^+} h(x)\right) \left( \lim_{x \to 1^+} g(x) \right) = (3)(2) = 6$

Because $\displaystyle \lim_{x \to 1^-} \left[ h(x)g(x) \right] \neq \lim_{x \to 1^+} \left[ h(x)g(x) \right]$, we know $\displaystyle \lim_{x \to 1} h(x)g(x)$ does not exist.

The definition of continuity at $x=1$ is $\displaystyle h(1)g(1) = \lim_{x \to 1}\left[ h(x)g(x) \right]$. Since the limit does not exist, the function is not continuous regardless of the value given for the constant.