# Solving Limits: A Shortcut to Finding Limits Using the Conjugate Theorem

In summary: I think you have to check the conditions first, and then see the theorem as a tool to get a result. So it is not about conditions, but about a result. By the way. What is the result?
<Moderator's note: Moved from a technical forum and thus no template.>

How to find this limit?

$$\lim_{x \to 0} \frac{5x} {3 -\sqrt{9-x}}$$

I'd tried to find this limit as below but the result is 0:

$$\lim_{x \to 0} \frac{5x} {3 -\sqrt{9-x}}$$
$$\lim_{x \to 0} \frac{5x} {3 - \sqrt{9-x}} × \frac{3 + \sqrt{9 - x}}{3 + \sqrt{9 - x}}$$
$$\lim_{x \to 0} \frac{5x (3 + \sqrt{9 - x})} {(3 -\sqrt{9-x})(3 + \sqrt{9 - x})}$$
$$\lim_{x \to 0} \frac{15x + 5x\sqrt{9 - x}} {9 - (9 - x)}$$
$$\lim_{x \to 0} \frac{15x + \sqrt{5x}\sqrt{9 - x}} {9 - (9 - x)}$$

Last edited by a moderator:
How to find this limit?

$$\lim_{x \to 0} \frac{5x} {3 -\sqrt{9-x}}$$

I'd tried to find this limit as below but the result is 0:

$$\lim_{x \to 0} \frac{5x} {3 -\sqrt{9-x}}$$
$$\lim_{x \to 0} \frac{5x} {3 - \sqrt{9-x}} × \frac{3 + \sqrt{9 - x}}{3 + \sqrt{9 - x}}$$
$$\lim_{x \to 0} \frac{5x (3 + \sqrt{9 - x})} {(3 -\sqrt{9-x})(3 + \sqrt{9 - x})}$$
$$\lim_{x \to 0} \frac{15x + 5x\sqrt{9 - x}} {9 - (9 - x)}$$
So far, so good. Why did you stop here (and inserted a mistake)?
$$\lim_{x \to 0} \frac{15x + \sqrt{5x}\sqrt{9 - x}} {9 - (9 - x)}$$

I prefer L' hopital rule in such cases.
The given question is 0/0 indeterminate form , so just apply it and you will get answer ( which is 30 , right? ) without even touching your pen .

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AgNO3 said:
I prefer L' hopital rule in such cases.
Except that is often the case that L'Hopital's Rule is no help, such as when you apply it a couple of times and end up with basically the same limit you started with. For problems like this one, I always start by working with the conjugate, as the OP was doing.

Mark44 said:
Except that is often the case that L'Hopital's Rule is no help, such as when you apply it a couple of times and end up with basically the same limit you started with. For problems like this one, I always start by working with the conjugate, as the OP was doing.

True.
But in many cases it either gets highly simplified or less complex.
Every question has its own easy approach, and i have enough practice on this topic to "see" through these questions.
I used it in this particular question bcoz it gave away answer with very simple mental calculation. So why should we waste time in working with the conjugates ?

I prefer to expand the root term. With appropriate cancelling, quickly get y = 30/(1-x/36...)

Edit: I'm wrong

Last edited:
atomicpedals said:
Between these two steps I think you have a sign error in multiplying out the denominator (FOIL-ing).
$$(3 - \sqrt{9 - x}) (3 + \sqrt{9 - x}) \neq {9 - (9 - x)}$$
Edit: For clarity

Let me write in detail

##(3 - \sqrt{9 - x}) (3 + \sqrt{9 - x})##
##= (3 × 3) - (3 × \sqrt{9 - x}) + (3 × \sqrt{9 - x}) - (\sqrt{9 - x})^2##
##= 9 - (9 - x)##
##= 9 - 9 + x)##
##= 0 + x##
##= x##

So, why do you said ##(3 - \sqrt{9 - x}) (3 + \sqrt{9 - x}) \neq {9 - (9 - x)}##?

Let me write in detail

##(3 - \sqrt{9 - x}) (3 + \sqrt{9 - x})##
##= (3 × 3) - (3 × \sqrt{9 - x}) + (3 × \sqrt{9 - x}) - (\sqrt{9 - x})^2##
##= 9 - (9 - x)##
##= 9 - 9 + x)##
##= 0 + x##
##= x##

So, why do you said ##(3 - \sqrt{9 - x}) (3 + \sqrt{9 - x}) \neq {9 - (9 - x)}##?
Because I'm a fool and completely missed the parenthesis in your math! If you take it down to x, then the problem becomes much easier (most of your x's cancel).

How do I cancel the x? Please write it in latex.

What you're left with is
$$\frac {15x + 5x \sqrt{9-x}} {x}=\frac {15x} {x} + \frac {5x \sqrt{9-x}} {x}$$

$$\lim_{x \to 0} \frac{5x} {3 -\sqrt{9-x}}$$
$$\lim_{x \to 0} \frac{5x} {3 - \sqrt{9-x}} × \frac{3 + \sqrt{9 - x}}{3 + \sqrt{9 - x}}$$
$$\lim_{x \to 0} \frac{5x (3 + \sqrt{9 - x})} {(3 -\sqrt{9-x})(3 + \sqrt{9 - x})}$$
$$\lim_{x \to 0} \frac{15x + 5x\sqrt{9 - x}} {9 - (9 - x)}$$
That is where you stopped. Now just go ahead:
How do I cancel the x? Please write it in latex.
##=\lim_{x \to 0} \dfrac{x\cdot (15+5\sqrt{9-x})}{9-9+x}=\lim_{x \to 0} \left(\dfrac{x}{x}\cdot (15+5\sqrt{9-x})\right)##

AgNO3 said:
But in many cases it either gets highly simplified or less complex.
L'Hopital's is very often unsuccessful when there is a radical in either the numerator or denominator. After you apply this method twice, you get back to the same expression you started with, which is no help at all.

AgNO3 said:
Every question has its own easy approach, and i have enough practice on this topic to "see" through these questions.
Maybe, or maybe not. Let me know after you've taught this stuff for a few years.

Mark44 said:
Except that is often the case that L'Hopital's Rule is no help, such as when you apply it a couple of times and end up with basically the same limit you started with. For problems like this one, I always start by working with the conjugate, as the OP was doing.
I strongly agree with this. The (by me) suspected meaning of the exercise is less to solve a specific problem, but rather to see patterns and learn techniques. An expression ##a-\sqrt{b}## in the denominator longs for the conjugate. ##(a-b)(a+b)=a^2-b^2## is of such great importance in so many areas, that it is necessary to establish an automatism - regardless of the specific problem. IMO this is what this exercise is about. Furthermore I would not take out the big cannon in cases where the cancellation of a common factor is already a problem. To me such a solution doesn't help here and serves quite a different purpose, which is not to help out. The OP almost solved the problem, so there is no reason for a restart with heavy machinery, the more as whenever one uses a theorem one has to check the conditions first, and cannot simply apply it. And I haven't seen, that the preliminaries had been checked anywhere.

Mark44 said:
Maybe, or maybe not. Let me know after you've taught this stuff for a few years.

Yeah sure !
Apparently i haven't seen much cases in math where it gave me trouble . And when i was taught this theorem i was really excited coz before that, my teacher forced me to solve by expanding and stuff .

## 1. How do I find the limit using algebraic manipulation?

To find the limit of a function using algebraic manipulation, you can try simplifying the function by factoring, canceling out common terms, or using other algebraic techniques. Once you have simplified the function, you can plug in the limit value to find the limit.

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

A one-sided limit only considers the behavior of the function as it approaches the limit value from one direction (either the left or the right). A two-sided limit considers the behavior of the function as it approaches the limit value from both directions.

## 3. When should I use the squeeze theorem to find a limit?

The squeeze theorem is useful when you have a function that is sandwiched between two other functions whose limits are known. By comparing the values of these two functions, you can determine the limit of the middle function.

## 4. Can I use L'Hopital's rule to find any limit?

No, L'Hopital's rule can only be used to find the limit of a function if the limit results in an indeterminate form (such as 0/0 or ∞/∞). It cannot be used to find limits that result in other values.

## 5. How do I know if a limit does not exist?

A limit does not exist if the function approaches different values from the left and right sides of the limit value, or if the function approaches ∞ or -∞ as the limit value is approached. Additionally, a limit may not exist if the function has a jump or discontinuity at the limit value.

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