Finding the Expression for a Limit of 0 Using Derivative Definition

[ussjt]

If https://webwork2.math.ohio-state.edu/courses/math151wi06sm/tmp/png/Hmwk3/391484/kopko.4-prob17image1.png , find [PLAIN]https://webwork2.math.ohio-state.edu/courses/math151wi06sm/tmp/png/Hmwk3/391484/kopko.4-prob17image2.png using the definition of derivative. [PLAIN]https://webwork2.math.ohio-state.edu/courses/math151wi06sm/tmp/png/Hmwk3/391484/kopko.4-prob17image2.png is the limit as https://webwork2.math.ohio-state.edu/courses/math151wi06sm/tmp/png/Hmwk3/391484/kopko.4-prob17image4.png of the expression (answer)
I need help with finding the expression.

The value of the limit is 0.

Any help would be great.

 

[Hurkyl]

What have you done?

 

[VietDao29]

f'(x₀) is defined to be
$$f'(x_0) = \lim_{h \rightarrow 0} \frac{f(x_0 + h) - f(x_0)}{h}$$
Apply that to your function:
$$f'(9) = \lim_{h \rightarrow 0} \frac{f(9 + h) - f(9)}{h}$$. Now, how can you do this limit?

 

[ussjt]

I am just confused with the 1/2^.5 part.

I am used to seeing something like f(x)= 1/x...so that you make it 1/(9+h).

 

[VietDao29]

$$f(x) := \frac{1}{\sqrt{2}}$$ means that this function is a constant and does not depend on what x you choose.
It will return the same value (i.e, 1 / (2^.5)), no matter what x is. For example:
If $$g(x) := \frac{1}{\sqrt{3}}$$, then:
$$g(0) = \frac{1}{\sqrt{3}}$$,
$$g(7) = \frac{1}{\sqrt{3}}$$,
$$g(9) = \frac{1}{\sqrt{3}}$$,...
Do you understand? Can you solve the problem now?

 

[Jameson]

And if the limit definition doesn't click, look at it graphically. A constant function is just a horizontal line, which obviously has a slope of 0.

 

[ussjt]

I have looked at it graphically, but I having a hard time understanding how to make it into an expression...lim h->0 something.

thanks for all of the input, but I am still confused.

 

[Jameson]

Ok. VietDao29 showed you the limit definition of a derivative. It's the following...

$$f'(x)=\lim_{h\rightarrow{0}} \frac{f(x+h)-f(x)}{h}$$

Your function is $$\frac{1}{\sqrt{2}}$$, so I'll start you out.

$$f'(9)=\lim_{h\rightarrow{0}} \frac{f(9+h)-f(9)}{h}$$

This is still from VietDao29's post.

What is f(9)? What is f(0)? What is f of anything? If you can answer that, you can do your limit.

 

[VietDao29]

Another hint is that:
$$\frac{f(9 + h) - f(9)}{h} = \frac{\frac{1}{\sqrt 2} - \frac{1}{\sqrt 2}}{h} = 0, \ \forall h \neq 0$$
What does the above expression tell you?

 

[ussjt]

thanks, I got it now. Thanks a lot!