Relating limits to derivatives, as x approaches non zero number?

In summary, the conversation discussed finding the limit as x approaches 2 of [f(x)-f(2)]/[sqrt(x)-sqrt(2)], given that f'(2)=3. After attempting to solve the problem using the derivative definition, it was suggested to rationalize the denominator and think about the derivatives of f(2) and sqrt(x) at x=2. Using this approach, it was determined that the limit is equal to 6*sqrt(2). The conversation also discussed using LaTeX to format mathematical symbols and equations.
  • #1
i_m_mimi
17
0

Homework Statement



Suppose that f' (2) = 3. Find the limit as x approaches 2 of [f(x)−f(2)]/[sqrt(x) - sqrt(2)]

Answer: 6*sqrt 2

Homework Equations



The Attempt at a Solution



f'(x) = lim h->0 = [f(a +h) - f(a)]/h = slope [f(x)-f(2)]/ x-2

a = 2

i would think that the limit = 3? wrong answer

tried graphing it, but no original function is given
sqrtx and sqrt2 confuses me and not sure how to do it.

thank you
 
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  • #2
i_m_mimi said:

Homework Statement



Suppose that f' (2) = 3. Find the limit as x approaches 2 of [f(x)−f(2)]/[sqrt(x) - sqrt(2)]

Answer: 6*sqrt 2

Homework Equations



The Attempt at a Solution



f'(x) = lim h->0 = [f(a +h) - f(a)]/h = slope [f(x)-f(2)]/ x-2

a = 2

i would think that the limit = 3? wrong answer

tried graphing it, but no original function is given
sqrtx and sqrt2 confuses me and not sure how to do it.

thank you
Rationalize the denominator of ##\displaystyle \ \frac{f(x)−f(2)}{\sqrt{x\,}-\sqrt{2\,}}\ .##
 
  • #3
Equivalently, write
[tex]\frac{f(x)- f(2)}{\sqrt{x}- \sqrt{2}}= \frac{f(x)- f(2)}{x- 2}\frac{x- 2}{\sqrt{x}- \sqrt{2}}[/tex]
[tex]= \frac{f(x)- f(2)}{x- 2}\left(\frac{1}{\frac{\sqrt{x}- \sqrt{2}}{x- 2}}\right)

\text{ and think about the derivatives of f(2) and }\sqrt{x}[/tex] at x= 2.
 
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  • #4
HallsofIvy said:
Equivalently, write
[tex]\frac{f(x)- f(2)}{\sqrt{x}- \sqrt{2}}= \frac{f(x)- f(2)}{x- 2}\frac{x- 2}{\sqrt{x}- \sqrt{2}}[/tex]
[tex]= \frac{f(x)- f(2)}{x- 2}\left(\frac{1}{\frac{\sqrt{x}- \sqrt{2}}{x- 2}}\right)

\text{ and think about the derivatives of f(2) and }\sqrt{x}[/tex] at x= 2.

I think I've got it :) how did you create those math symbols and format it that way? was it through the physicsforums website coding or by a program?


f'(2) = 3
x = 2

lim x->2 [f(x) - f(2)]/(x - 2) * lim x->2 1/ [sqrtx - sqrt2]/(x-2)

= (3) * 1/f'(sqrtx)
= 3 * 1/(1/2sqrtx)
= 3 * 2sqrtx
= 3 * 2 sqrt2
= 6 sqrt2
 
  • #5
It's called ##\LaTeX## ;)

Underneath each post you will see a button marked "quote" - if you click on it, you will be taken to an advanced editor screen where everything in the post is included in "quote" tags. Read what is in the quote to see how all the symbols came out nicely formatted ;)

Use double-hash or double-dollar signs around the block of math you want formatted - and use a backslash "\" in front of the bits that need special formatting - so \sqrt{ \frac{\alpha}{\beta} } will come out $$\sqrt{\frac{\alpha}{\beta}}$$
 

Related to Relating limits to derivatives, as x approaches non zero number?

1. What is the definition of a limit?

A limit is a mathematical concept that represents the value a function approaches as the input approaches a certain value, often denoted by the symbol "x" or "a". It is used to describe the behavior of a function near a specific point.

2. How is a limit related to derivatives?

The derivative of a function at a point is defined as the slope of the tangent line to the graph of the function at that point. This slope can also be thought of as the rate of change of the function at that point. The limit of the function as the input approaches that point is equal to the derivative of the function at that point.

3. Can a function have a limit but no derivative?

Yes, it is possible for a function to have a limit at a point but not have a derivative at that point. This occurs when the function is not continuous at that point, meaning that there is a gap or jump in the graph of the function at that point.

4. How do you find the limit of a function as x approaches a non-zero number?

To find the limit of a function as x approaches a non-zero number, you can simply plug in the value of the number into the function and evaluate. If the resulting value is undefined, you can use algebraic techniques such as factoring or rationalizing the denominator to simplify the expression and find the limit.

5. Why is it important to understand the relationship between limits and derivatives?

The relationship between limits and derivatives is crucial in understanding the behavior of a function near a specific point. It allows us to determine the slope or rate of change of a function at a point, which has many real-world applications in fields such as physics, engineering, and economics. Additionally, it provides a foundation for more advanced concepts in calculus and other branches of mathematics.

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