What is the correct function f(x) to use for this limit?

The correct function is f(x) = 1/(3√x-9).In summary, the conversation is about solving a limit question involving a function and using substitution to simplify the equation. The person tried using a function but it was not the correct one. The correct function is f(x) = 1/(3√x-9).
  • #1
subasurf
1
0
Hi guys, I'm really new to calculus and limits and have been trying to have a good crack at the following question. Sorry if I haven't written the problem out in the most acceptable format.

lim (9-3√x)/(9-x)
x→9

Substituting 9 gives you 0/0 and indeterminate.

I tried multiplying the numerator and denominator by (-9+3√x), and multiplied out the brackets to wind up with -81/-27 and then simplified down to - 3/1

I'm just not sure if I'm going about this the correct way. I don't necessarily want the answer, just a bit of guidance. Thanks guys.
 
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  • #2
subasurf said:
I tried multiplying the numerator and denominator by (-9+3√x), and multiplied out the brackets to wind up with -81/-27 and then simplified down to - 3/1

There is some function f(x) that you can multiply numerator and denominator with, with the property that (9-3√x)f(x) = 9-x, but the function you are suggesting above is not it.
 

Related to What is the correct function f(x) to use for this limit?

1. What is a limit with square roots?

A limit with square roots is a mathematical concept that describes the behavior of a function as the input gets closer and closer to a specific value, typically represented by the variable x. It is denoted by the notation "lim x→a f(x)", where "f(x)" is the function and "a" is the specific value that the input is approaching.

2. How do you evaluate a limit with square roots?

To evaluate a limit with square roots, you can use the properties of limits, such as the limit laws, to simplify the expression. You can also use algebraic manipulations, such as multiplying by the conjugate, to eliminate square roots within the expression. Once the expression is simplified, you can substitute the specific value for the variable and solve for the limit.

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

A one-sided limit with square roots only considers the behavior of the function as the input approaches the specific value from one direction, either the left or the right. A two-sided limit with square roots considers both directions and requires that the function approaches the same value from both sides in order for the limit to exist.

4. Can a limit with square roots have a different value from the actual value of the function?

Yes, a limit with square roots can have a different value from the actual value of the function. This can occur when the function is not defined at the specific value or when there is a discontinuity in the function at that point. In these cases, the limit represents the behavior of the function as the input gets closer to the specific value, but it may not equal the actual value of the function at that point.

5. How can limits with square roots be used in real-life applications?

Limits with square roots can be used in various real-life applications, such as in physics and engineering, to model and predict the behavior of a system as a variable approaches a certain value. For example, they can be used to calculate the maximum velocity of a falling object or the maximum load a bridge can support. They are also used in finance and economics to analyze trends and predict future values.

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