How do evaluate this limit because i will get 2.6667 and divide by -144

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In summary, the conversation discusses evaluating a limit and plugging it into a tangent line equation. The correct answer is given, but the person is unsure of how it was obtained and asks for clarification on the definitions being used. It is explained that the two definitions given are equivalent and the person is encouraged to continue working on the problem.
  • #1
Teh
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My Work:

View attachment 6111

How do evaluate this limit because i will get 2.6667 and divide by -144 it will get me 55.38446...
I know that the answer is WRONG but I can't figure it out.
Then after that i have to plug in into tangent line y-f(a) = Mtan(x-a)

what i am doing wrong.

The Problem:
View attachment 6110
 

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  • #2
Teh said:
How do evaluate this limit because i will get 2.6667 and divide by -144 it will get me 55.38446...
I know that the answer is WRONG but I can't figure it out.
Then after that i have to plug in into tangent line y-f(a) = Mtan(x-a)

what i am doing wrong.
The problem:

Do we have at the denominator $\boldsymbol + 4+12 \sqrt{x}$ ? If so then by substitung at $x$ the value $\frac{1}{9}$ and since $\sqrt{9}=3$, the limit is equal to $\frac{-144}{\frac{1}{3}\left( 4+\frac{12}{3}\right)}=\frac{-144 \cdot 3}{\left( 4+\frac{12}{3}\right)}=\frac{-144 \cdot 3}{\frac{24}{3}}=\frac{-144 \cdot 9}{24}=-6 \cdot 9=-54$.

Also it holds that $\left( \frac{1}{\sqrt{x}}\right)'=-\frac{1}{2 x^{\frac{3}{2}}}$.
 
  • #3
Because you speak of limits, I am assuming you are to use the definition:

\(\displaystyle f'(x)\equiv\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\)

where:

\(\displaystyle f(x)=\frac{4}{\sqrt{x}}\)

So, let's look at what we'll be taking the limit of:

\(\displaystyle \frac{\dfrac{4}{\sqrt{x+h}}-\dfrac{4}{\sqrt{x}}}{h}=4\frac{\sqrt{x}-\sqrt{x+h}}{h\sqrt{x(x+h)}}=-\frac{4}{\sqrt{(x(x+h))}\left(\sqrt{x}+\sqrt{x+h}\right)}\)

So, we now have a determinate form, and we can state:

\(\displaystyle f'(x)=-4\lim_{h\to0}\frac{1}{\sqrt{(x(x+h))}\left(\sqrt{x}+\sqrt{x+h}\right)}\)

Can you proceed to determine the limit now?
 
  • #4
What the..HOW lol dang it ahhhh... -.- let me see what you did

- - - Updated - - -

MarkFL said:
Because you speak of limits, I am assuming you are to use the definition:

\(\displaystyle f'(x)\equiv\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\)

where:

\(\displaystyle f(x)=\frac{4}{\sqrt{x}}\)

So, let's look at what we'll be taking the limit of:

\(\displaystyle \frac{\dfrac{4}{\sqrt{x+h}}-\dfrac{4}{\sqrt{x}}}{h}=4\frac{\sqrt{x}-\sqrt{x+h}}{h\sqrt{x(x+h)}}=-\frac{4}{\sqrt{(x(x+h))}\left(\sqrt{x}+\sqrt{x+h}\right)}\)

So, we now have a determinate form, and we can state:

\(\displaystyle f'(x)=-4\lim_{h\to0}\frac{1}{\sqrt{(x(x+h))}\left(\sqrt{x}+\sqrt{x+h}\right)}\)

Can you proceed to determine the limit now?

for my problem i won't be using f(a+h) - f(h) instead Ill be using this...View attachment 6112but is there any different?
 

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  • #5
These are two different definitions. Both give the same result.
 
  • #6
evinda said:
These are two different definitions. Both give the same result.

ahh...okay thank you very much
 
  • #7
What I gave and the example you have been given use different notation, but they are equivalent. They are using:

\(\displaystyle a=x+h\)

So, as $h\to0$ we have $a\to x$. They are telling you to use:

\(\displaystyle f'(x)\equiv\lim_{x\to a}\frac{f(x)-f(a)}{x-a}\)

So, using the given definition of $f$, what we are taking the limit of is:

\(\displaystyle \frac{\dfrac{4}{\sqrt{x}}-\dfrac{4}{\sqrt{a}}}{x-a}=\frac{4\left(\sqrt{a}-\sqrt{x}\right)}{(x-a)\sqrt{ax}}=-\frac{4}{\sqrt{ax}(\sqrt{a}+\sqrt{x})}\)

And so you now have the determinate form:

\(\displaystyle f'(x)=-4\lim_{x\to a}\frac{1}{\sqrt{ax}(\sqrt{a}+\sqrt{x})}\)
 

1. What is a limit in mathematics?

A limit is a fundamental concept in mathematics that describes the behavior of a function as its input approaches a certain value. It is used to determine the value that a function "approaches" as the input gets closer and closer to a specific value, without ever reaching it.

2. How do you evaluate a limit?

To evaluate a limit, you first substitute the value that the input is approaching into the function. Then, you simplify the resulting expression as much as possible, and if necessary, use algebraic techniques or mathematical rules (such as L'Hôpital's rule) to further simplify the expression. Finally, you plug in the value of the input to get the limit's numerical value.

3. What is the difference between a one-sided limit and a two-sided limit?

A one-sided limit only considers the behavior of a function as the input approaches a specific value from one side (either the left or the right). A two-sided limit, on the other hand, considers the behavior of the function as the input approaches the specific value from both the left and the right. In other words, a one-sided limit only looks at the behavior of a function as the input approaches the value "from one direction", while a two-sided limit looks at the behavior from both directions.

4. Can a limit exist even if the function is not defined at that point?

Yes, a limit can exist even if the function is not defined at that point. This is because a limit is only concerned with the behavior of a function as the input approaches a specific value, not the actual value of the function at that point. Therefore, even if the function is not defined at that point, the limit can still exist.

5. What are some common techniques for evaluating limits?

Some common techniques for evaluating limits include direct substitution, factoring, finding common denominators, and using algebraic manipulation. In more advanced cases, techniques such as L'Hôpital's rule, trigonometric identities, and Taylor series may also be used.

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