# Thread: Differentiation and Need HELP with LATEX!

1. ## Differentiation and Need HELP with LATEX!

g(x) = {-x, x<=0
3x^2,x>0

(a) Evaluate the limit of {g(x+delta x) - g(x)}/{delta x) for x<=0 and x>0 as delta x tends to 0

(b) evaluate the limit of {g(delta x) - g(0)}/{delta x) as delta x tends to 0 from the right and as delta x tends to 0 from the left

(c) sketch the graph of g'(x)

Is g(x) continuous at x=0?

I need help with (b) and (c) only. Is there any user-friendly programmes to type the equations? I find the syntax of latex very difficult to learn. Btw, how do you type the above equations (using latex)?

Thanks!

2. B is simply asking you to compute the separate limits for the left and right side of the equation corresponding the respective piece of your piece-wise function. You are meant to see whether the limit from the right is the same as the limit from the left.

3. Originally Posted by ANDS!
B is simply asking you to compute the separate limits for the left and right side of the equation corresponding the respective piece of your piece-wise function. You are meant to see whether the limit from the right is the same as the limit from the left.
Does this mean I must find the limit for $-x$, $x\leq 0$ as
delta x approaches 0 from the right and from the left as well as the limit for $3x^2$, $x>0$ as delta x approaches 0 from the right and from the left?

Can anyone show me the working??? Thank you.

4. Originally Posted by cyt91
Does this mean I must find the limit for $-x$, $x\leq 0$ as
delta x approaches 0 from the right and from the left as well as the limit for $3x^2$, $x>0$ as delta x approaches 0 from the right and from the left?

Can anyone show me the working??? Thank you.
For (a), you want to take x< 0 (not $\le 0$) and can assume that delta x is small enough that x+ delta x is also negative- then you can use "-x" as the formula for both g(x) and g(x+ delta x). If g(x)= -x, what is g(x+ delta x)?

Then take x> 0 and assume that delta x is small enough that x+ delta x is also positive- then you can use " $3x^2$ as the formula for both g(x) and g(x+ delta x). If $g(x)= 3x^2$, what is g(x+ deltax)?

For (b), where x=0, taking the limit "from the left" means using g(0+ delta x)= -delta x and taking the limit "from the right" means using $g(0+ delta x)= 3(delta x)^2$.

5. Originally Posted by HallsofIvy
For (a), you want to take x< 0 (not $\le 0$) and can assume that delta x is small enough that x+ delta x is also negative- then you can use "-x" as the formula for both g(x) and g(x+ delta x). If g(x)= -x, what is g(x+ delta x)?

Then take x> 0 and assume that delta x is small enough that x+ delta x is also positive- then you can use " $3x^2$ as the formula for both g(x) and g(x+ delta x). If $g(x)= 3x^2$, what is g(x+ deltax)?

For (b), where x=0, taking the limit "from the left" means using g(0+ delta x)= -delta x and taking the limit "from the right" means using $g(0+ delta x)= 3(delta x)^2$.
For (a) :
x<0,
g(x+delta x)= -(x+delta x)
Is this correct?
Then, the limit for [g(x+delta x)-g(x)]/[delta x] as delta x tends to 0 is
{-(x+delta x)-[-x]}/{delta x}=-1
Correct?

x>0,
g(x+delta x)=3(x+delta x)^2
Is this correct?
Then the limit for [g(x+delta x)-g(x)]/[delta x] as delta x tends to 0 is
{3(x+delta x)^2-3x^2}/[delta x]=6x
Correct?

For (b):
when delta x approaches 0 from the left, the limit of
{g(0+delta x)-g(0)}/{delta x} = {-(0+delta x)-[-0]}/[delta x]
=-1

and when delta x approaches 0 from the right, the limit of
{g(0+delta x)-g(0)}/{delta x}
= {3(0+delta x)^2-3(delta x)^2}/[delta x]
=0

Are my workings correct?

How about the graph? Is it g'(x)=-1 for x<=0 and g'(x)=6x for x>0?

Hence, is g(x) not continuous at x=0 since g'(x) for x<=0 and g'(x) for x>0 are not equal?

Thank you!

6. Originally Posted by cyt91
For (a) :
x<0,
g(x+delta x)= -(x+delta x)
Is this correct?
Then, the limit for [g(x+delta x)-g(x)]/[delta x] as delta x tends to 0 is
{-(x+delta x)-[-x]}/{delta x}=-1
Correct?
Yes, if x< 0 and $|\delta x|< |x|$, which we can assume since we are taking the limit as $\delta x$ goes to 0, $x+ \delta x$ is also < 0 so $f(x+ \delta x)- f(x)= -(x+\delta x)- x= -\delta x$.

x>0,
g(x+delta x)=3(x+delta x)^2
Is this correct?
Yes, if x> 0 and $\delta x> x$, which we can assume since we are taking the limit as $\delta x$ goes to 0, $x+ \delta x$ is also > 0 so $f(x+ \delta x)- f(x)= 3(x+ \delta x)^2- 3x^2$.

[quote]Then the limit for [g(x+delta x)-g(x)]/[delta x] as delta x tends to 0 is
{3(x+delta x)^2-3x^2}/[delta x]=6x
Correct?[/tex]
Yes. 3(x+ \delta x)^2= 3(x^2+ 2x\delta x+ \delta x^2)= 3x^2- 6x\delta x+ \delta x^2[/tex]. $3(x+ \delta x)^2- 3x^2= 6x\delta+ \delta x^2$ so $(3(x+ \delta x)^2- 3x^2)/\delta x= 6x+ \delta x$ and the limit of that, as $\delta x$ goes to 0 is 6x.

Very good!

For (b):
when delta x approaches 0 from the left, the limit of
{g(0+delta x)-g(0)}/{delta x} = {-(0+delta x)-[-0]}/[delta x]
=-1

and when delta x approaches 0 from the right, the limit of
{g(0+delta x)-g(0)}/{delta x}
= {3(0+delta x)^2-3(delta x)^2}/[delta x]
=0
Both correct.

Are my workings correct?

How about the graph? Is it g'(x)=-1 for x<=0 and g'(x)=6x for x>0?

Hence, is g(x) not continuous at x=0 since g'(x) for x<=0 and g'(x) for x>0 are not equal?

Thank you!
No, "continuous at x= a" mean $\lim_{x\to a} g(x)= g(a)$. It has nothing to do with g' because most continuous functions are not differentiable! And, in fact, to have a derivative, a function must be continuous.

Here, the point is that $\lim_{x\to 0^-} g(x)= \lim_{x\to 0} -x= 0$ and $\lim_{x\to 0^+} g(x)= \lim_{x\to 0} 3x^2= 0$.

So $\lim_{x\to 0} g(x)= 0= g(0)$ and g is continuous at x= 0.