Continuity of Piecewise Function

[Jason76]

\(\displaystyle h(x) = \dfrac{x^{2} + x}{x}\) if \(\displaystyle x\) does NOT \(\displaystyle = 0\)

\(\displaystyle h(x) = 1\) if \(\displaystyle x = 0\)

Show whether \(\displaystyle h\) is continuous at \(\displaystyle x = 0\). Use 3 steps.

Any starting hint?

 

[Jason76]

Should we look at the left and right limits.?

 

[stapel]

\(\displaystyle h(x) = \dfrac{x^{2} + x}{x}\) if \(\displaystyle x\) does NOT \(\displaystyle = 0\)

\(\displaystyle h(x) = 1\) if \(\displaystyle x = 0\)

Show whether \(\displaystyle h\) is continuous at \(\displaystyle x = 0\). Use 3 steps.

Any starting hint?

I'd start with finding the limit of the first rule, as x approaches zero. I have no idea what "three steps" they're expecting you to use, though.

 

[Jason76]

The limit of the top function would start out as indeterminate \(\displaystyle \dfrac{0}{0}\)

So you would use L Hopitals rule (or factoring and canceling out) and get \(\displaystyle 1\) as the limit.

Since both the top and bottom functions have limits of \(\displaystyle 1\). Then perhaps it's continuous, but I don't know anything about any 3 steps.

 

[JeffM]

The limit of the top function would start out as indeterminate \(\displaystyle \dfrac{0}{0}\)

So you would use L Hopitals rule (or factoring and canceling out) and get \(\displaystyle 1\) as the limit.

Since both the top and bottom functions have limits of \(\displaystyle 1\). Then perhaps it's continuous, but I don't know anything about any 3 steps.

I suspect that they are going back to the definition of continuous

\(\displaystyle f(x)\ is\ continuous\ at\ a\ if\ and\ only\ if\)

\(\displaystyle (1)\ f(a) \in \mathbb R;\ and\)

\(\displaystyle (2)\ \displaystyle \lim_{x \rightarrow a^+}f(x) = f(a);\ and\)

\(\displaystyle (3)\ \displaystyle \lim_{x \rightarrow a^-}f(x) = f(a).\)

Edit: By the way, you do not need L'Hospital's Rule at all. \(\displaystyle x \ne 0 \implies f(x) = \dfrac{x^2 + x}{x} = x + 1\) is all you need.

 

[HallsofIvy]

I would NOT say that "the limit starts out as indeterminant [itex]\frac{0}{0}[/itex]". An "inderminant" is not a number so it really makes no sense to say that a limit (which is a number) "is" or "starts out as" an indeterminant.

Also, I, personally, would not use something as "sophisticated" as L'Hopital's rule. I would note that, as long as x is not equal to 0, \(\displaystyle \frac{x^2+ x}{x}= \frac{x(x+ 1)}{x}= x+ 1\). Now take the limit, as x goes to 0, of x+ 1.

(One of the properties of limits you should have learned is
"If f(x)= g(x) for all x except x= a, then \(\displaystyle \lim_{x\to a} f(x)= \lim_{x\to a} g(x)\)."

 

[HallsofIvy]

I would NOT say that "the limit starts out as indeterminant [itex]\frac{0}{0}[/itex]". An "inderminant" is not a number so it really makes no sense to say that a limit (which is a number) "is" or "starts out as" an indeterminant.

Also, I, personally, would not use something as "sophisticated" as L'Hopital's rule. I would note that, as long as x is not equal to 0, \(\displaystyle \frac{x^2+ x}{x}= \frac{x(x+ 1)}{x}= x+ 1\). Now take the limit, as x goes to 0, of x+ 1.

(One of the properties of limits you should have learned is
"If f(x)= g(x) for all x except x= a, then \(\displaystyle \lim_{x\to a} f(x)= \lim_{x\to a} g(x)\)."

Perhaps this is what is meant by the "three steps"-

A function, f, is continuous at x= a if and only if these three statements are true:
1) f(a) exists.

2) \(\displaystyle \lim_{x\to a} f(x)\)

3) \(\displaystyle \lim_{x\to a} f(x)= f(a)\)

Here, (1) is true because h(0) is defined as 1. A good way to show that (2) is true is to actually find the limit. Finally, if that limit is 1, then (3) is true.

 

[Jason76]

Three steps:

1. \(\displaystyle f(a)\) must exist

In the case of asymptotic functions, \(\displaystyle f(a)\) would not exist.

2. \(\displaystyle \lim x \rightarrow a\) must exist (referring to total limit, not just the right or left ones).

The second step would be true if it had a limit (as with the case in graphs with a hole, or ones with no jumps, holes, or asymptotes). It would fail in the case of jumps as the right and left limits would not be equal.

3. \(\displaystyle \lim x \rightarrow a = f(a)\)

The third step would fail in the case of holes in the graph, as there would be a total limit, but it wouldn't equal the \(\displaystyle f(a)\).

 

[JeffM]

Three steps:

1. \(\displaystyle f(a)\) must exist

In the case of asymptotic functions, \(\displaystyle f(a)\) would not exist.

2. \(\displaystyle \lim x \rightarrow a\) must exist (referring to total limit, not just the right or left ones).

The second step would be true if it had a limit (as with the case in graphs with a break, or ones with no breaks, jumps, or asymptotes).

3. \(\displaystyle \lim x \rightarrow a = f(a)\)

The third step would fail in the case of holes in the graph, as there would be a total limit, but it wouldn't equal the \(\displaystyle f(a)\).

The limit, what you are calling the total limit, exists only if both the right and left limits exist and are equal.

 

[Jason76]

The limit, what you are calling the total limit, exists only if both the right and left limits exist and are equal.

Right, that's what I was getting at.

 

[Jason76]

\(\displaystyle h(x) = \dfrac{x^{2} + x}{x}\) if \(\displaystyle x\) does NOT \(\displaystyle = 0\)

\(\displaystyle h(x) = 1\) if \(\displaystyle x = 0\)

Show whether \(\displaystyle h\) is continuous at \(\displaystyle x = 0\). Use 3 steps.

This function is continuous as

\(\displaystyle \lim x \rightarrow 0- = \lim x \rightarrow 0+\), \(\displaystyle f(0)\) exists, and \(\displaystyle \lim x \rightarrow 0 = f(0)\)

Proof:

\(\displaystyle f(0) = 1\) - It exists.

\(\displaystyle \lim x \rightarrow 0- \dfrac{(0)^{2} + 0}{0} = \dfrac{0}{0}\)

Use L'Hopitals

\(\displaystyle \dfrac{2x + 1}{1} = \lim x \rightarrow 0- \dfrac{2(0) + 1}{1} = 1\)

\(\displaystyle \lim x \rightarrow 0+ = 1\)

\(\displaystyle 1 = 1\)

\(\displaystyle f(a) = 1\) and \(\displaystyle \lim x \rightarrow 0 = 1\)