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

[askor]

<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)}

 

[fresh_42]

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)?
Go ahead and cancel $x$.

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

 

[AgNO3]

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 .

 

[AgNO3]

 

[Mark44]

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.

 

[AgNO3]

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 ?

 

[mjc123]

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

 

[atomicpedals]

Edit: I'm wrong

 

[askor]

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 $(3 - \sqrt{9 - x}) (3 + \sqrt{9 - x}) \neq {9 - (9 - x)}$?

 

[atomicpedals]

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).

 

[askor]

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

 

[atomicpedals]

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

 

[fresh_42]

\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.

$=\lim_{x \to 0} \dfrac{x\cdot (15+5\sqrt{9-x})}{9-9+x}=\lim_{x \to 0} \left(\dfrac{x}{x}\cdot (15+5\sqrt{9-x})\right)$

 

[Mark44]

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.

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.

 

[fresh_42]

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.

 

[AgNO3]

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 .