# Thread: Using Composition With Continuity

1. ## Using Composition With Continuity

Let $g$ be defined on $\mathbb{R}$ by $g(1) := 0$, and $g(x) := 2$ if $x \neq 1$, and let $f(x) := x +1$ for all $x \in \mathbb{R}$. How would you show that $\lim_{x\to0}g \circ f \neq (g \circ f)(0)$ ?

2. Originally Posted by CrazyCat87
Let $g$ be defined on $\mathbb{R}$ by $g(1) := 0$, and $g(x) := 2$ if $x \neq 1$, and let $f(x) := x +1$ for all $x \in \mathbb{R}$. How would you show that $\lim_{x\to0}g \circ f \neq (g \circ f)(0)$ ?
$g(f(0))=g(0+1)=g(1)=0$ but $g(f(x))=g(x+1)=2,\text{ }x\ne 0$...

3. Originally Posted by Drexel28
$g(f(0))=g(0+1)=g(1)=0$ but $g(f(x))=g(x+1)=2,\text{ }x\ne 0$...
So to show they're not equal, should I prove that the limit of $\lim_{x\to0}g \circ f =2$ ?

4. Originally Posted by CrazyCat87
So to show they're not equal, should I prove that the limit of $\lim_{x\to0}g \circ f =2$ ?
Is there a need to prove it?

5. Originally Posted by Drexel28
Is there a need to prove it?
Yea, I'm trying to prove that $\lim_{x \to 0}g \circ f = 2$ so that I can show that the two are different, and I'm a bit stuck...

so $\lim_{x \to 0}g \circ f = \lim_{x \to 0}g(x+1)$

I wanna show $|g(x+1)-0|=|g(x+1)|<\epsilon$ ...

6. Originally Posted by CrazyCat87
Yea, I'm trying to prove that $\lim_{x \to 0}g \circ f = 2$ so that I can show that the two are different, and I'm a bit stuck...

so $\lim_{x \to 0}g \circ f = \lim_{x \to 0}g(x+1)$

I wanna show $|g(x+1)-0|=|g(x+1)|<\epsilon$ ...