Calculating Left Hand Limit: ##\sqrt 6##

[MatinSAR]

Homework Statement

Find the following limits.

Relevant Equations

Please see below...

$\lim_{x \rightarrow 3} {\frac { \sqrt {9-x^2} } {\sqrt x+\sqrt{3-x}-\sqrt3 }}$

I wanted to calculate left hand limit.
I find out that the answer is $\sqrt 6$ using GeoGebra.

What I have done:
I divided the numerator and denominator by $\sqrt {3-x}$.

$\lim_{x \rightarrow 3^-} {\frac { \sqrt {3+x} } {\sqrt {\frac {x} {3-x}}+1-\sqrt {\frac {3} {3-x}} }}=\frac {\sqrt 6} {\infty + 1 -\infty}=\sqrt 6$
I'm not sure If my method is right mathematically.

 

[anuttarasammyak]

The given formula equals
\lim_{x \rightarrow 3} \frac{\sqrt{3-x}\sqrt{3+x}}{\sqrt{3-x}}=\sqrt{6}
which is as same as yours. However, $\infty-\infty$ in your calculation is not good to use in general.

 

[MatinSAR]

The given formula equals
\lim_{x \rightarrow 3} \frac{\sqrt{3-x}\sqrt{3+x}}{\sqrt{3-x}}=\sqrt{6}
I got the same result.

Thank you for your reply. Final answer is true. We can find it using L'Hôpital's rule or using graphs.
I am not sure if my answer is right mathematically.

I am not sure if this part is correct:
$\frac {\sqrt 6} {\infty + 1 -\infty}=\sqrt 6$
And finally i'm not sure if $\lim_{x \rightarrow 3^-} {\frac {\sqrt {3+x}} 1}=\sqrt 6$.

 

[PeroK]

$\lim_{x \rightarrow 3^-} {\frac { \sqrt {3+x} } {\sqrt {\frac {x} {3-x}}+1-\sqrt {\frac {3} {3-x}} }}=\frac {\sqrt 6} {\infty + 1 -\infty}=\sqrt 6$
I'm not sure If my method is right mathematically.

It's not right at all. If $f(x) \rightarrow \infty$ and $g(x) \rightarrow \infty$, then $f(x) - g(x) \rightarrow$ could be anything.

What is correct in this case is to show separately that:
$$\lim_{x \rightarrow 3^-}\bigg ( \sqrt {\frac {x} {3-x}}-\sqrt {\frac {3} {3-x}}\bigg ) = 0$$

 

[MatinSAR]

It's not right at all. If $f(x) \rightarrow \infty$ and $g(x) \rightarrow \infty$, then $f(x) - g(x) \rightarrow$ could be anything.

What is correct in this case is to show separately that:
$$\lim_{x \rightarrow 3^-}\bigg ( \sqrt {\frac {x} {3-x}}-\sqrt {\frac {3} {3-x}}\bigg ) = 0$$

So, If I prove that limit is zero then my answer is right, isn't it?

 

[PeroK]

So, If I prove that the mentioned limit is zero then my answer is right, isn't it?

Yes. By calculating that limit, you see that both the numerator and denominator are well-defined at the limit point and the solution follows.

Let me show you a general idea in these cases. You have a quotient where the limit of the numerator is well defined and the limit of the denominator has an indeterminate form (in this case $\infty - \infty$).

If you calculate separately the limit of that indeterminate form, then (because the limit of the numerator and denominator both exist - this is an important point) you can combine them. In general:

If $\lim f(x)$ and $\lim g(x)$ both exist, then $\lim \frac{f(x)}{g(x)}$ exists and
$$lim \frac{f(x)}{g(x)} = \frac{\lim f(x)}{\lim g(x)}$$

 

[MatinSAR]

Yes. By calculating that limit, you see that both the numerator and denominator are well-defined at the limit point and the solution follows.

Let me show you a general idea in these cases. You have a quotient where the limit of the numerator is well defined and the limit of the denominator has an indeterminate form (in this case $\infty - \infty$).

If you calculate separately the limit of that indeterminate form, then (because the limit of the numerator and denominator both exist - this is an important point) you can combine them. In general:

If $\lim f(x)$ and $\lim g(x)$ both exist, then $\lim \frac{f(x)}{g(x)}$ exists and
$$lim \frac{f(x)}{g(x)} = \frac{\lim f(x)}{\lim g(x)}$$

Using your detailed guide, I have edited my answer. I hope it is right mathematically now...

 

[anuttarasammyak]

Addendum to my previous post :
The given formula is
\lim_{x \rightarrow 3-0} \frac{\sqrt{3-x}\sqrt{3+x}}{\frac{x-3}{\sqrt{x}+\sqrt{3}}+\sqrt{3-x}}
In the divider, the first term goes to zero with higher order than the second one so taking the limit we can disregard the first term to be zero.
I interpret that the limit as $\lim_{x \rightarrow 3-0}$ so that the relevant function is real.

 

[MatinSAR]

Addendum to my previous post :
The given formula is
\lim_{x \rightarrow 3-0} \frac{\sqrt{3-x}\sqrt{3+x}}{\frac{x-3}{\sqrt{x}+\sqrt{3}}+\sqrt{3-x}}
In the divider, the first term goes to zero with higher order than the second one so taking the limit we can disregard the first term to be zeo.
I interpret that the limit as $\lim_{x \rightarrow 3-0}$ so that the relevant function is real.

A clever idea. Thank you. I didn't undestand what's this one:

 

[anuttarasammyak]

Using your detailed guide, I have edited my answer. I hope it is right mathematically now...

\frac{\sqrt{x}-\sqrt{3}}{\sqrt{3-x}}=\frac{x-3}{\sqrt{3-x}(\sqrt{x}+\sqrt{3})}=-\frac{\sqrt{3-x}}{\sqrt{x}+\sqrt{3}} \rightarrow 0

 

[anuttarasammyak]

A clever idea. Thank you. I didn't undestand what's this one:

x is approaching to 3 from smaller side, e.g.
2.9, 2.99, 2.999, 2.999,...
Otherwise $\sqrt{3-x}$ would become imaginary.

 

[MatinSAR]

\frac{\sqrt{x}-\sqrt{3}}{\sqrt{3-x}}=\frac{x-3}{\sqrt{3-x}(\sqrt{x}+\sqrt{3})}=-\frac{\sqrt{3-x}}{\sqrt{x}+\sqrt{3}} \rightarrow 0

I understand. I've used L'Hôpital's rule but you have used $a^2-b^2=(a-b)(a+b)$.
So do you agree with my final answer?

 

[anuttarasammyak]

They coincide of cource and you are released from subtraction of infinities.

 

[MatinSAR]

They coincide of cource and you are released from subtraction of infinities.

Thank you for your help and time.

 

[MatinSAR]

@PeroK Thanks for your replies.

 

[Mark44]

However, ∞−∞ in your calculation is not good to use in general.

In fact, $[\infty - \infty]$ is one of several indeterminate forms, as already mentioned.

It's not right at all. If f(x)→∞ and g(x)→∞, then f(x)−g(x)→ could be anything.

Here are some examples that elaborate the point that @PeroK is making.

$\lim_{x \to \infty} x - x^2 = -\infty$
$\lim_{x \to \infty} x^3 - x^2 = \infty$
$\lim_{x \to \infty} (x + 2)^2 - (x^2 + 4x) = 4$

In each case, the expression we're taking the limit of is the indeterminate form $[\infty - \infty]$, but all three limits can be determined and all are different.

 

[vela]

Here's another way to solve the problem. Multiplying by a conjugate works in instances like this. Because $(\sqrt x - \sqrt 3)(\sqrt x + \sqrt 3) = x-3$, we can end up with a factor of $\sqrt{3-x}$ into every term so they will cancel out:
\begin{align*}\
\lim_{x \rightarrow 3^-} \frac { \sqrt {9-x^2} } {\sqrt x+\sqrt{3-x}-\sqrt3 }
&= \lim_{x \rightarrow 3^-} \frac { \sqrt {(3-x)(3+x)} } {\sqrt{3-x}-(\sqrt3-\sqrt x) } \cdot \frac{\sqrt{3} + \sqrt{x}}{\sqrt{3}+\sqrt{x}} \\
&= \lim_{x \rightarrow 3^-} \frac {(\sqrt{3} + \sqrt{x}) \sqrt {(3-x)(3+x)}}{(\sqrt{3} + \sqrt{x})\sqrt{3-x} -(3-x)} \\
&= \lim_{x \rightarrow 3^-} \frac {(\sqrt{3} + \sqrt{x}) \sqrt {3+x}}{\sqrt{3} + \sqrt{x} - \sqrt{3-x}}
\end{align*} Now there's no indeterminate form when you set $x=3$ so no need to use the Hospital rule to evaluate the limit. (Also, if you're allowed to use the Hospital rule, why not just use it right at the start?)

 

[MatinSAR]

In fact, $[\infty - \infty]$ is one of several indeterminate forms, as already mentioned.

Here are some examples that elaborate the point that @PeroK is making.

$\lim_{x \to \infty} x - x^2 = -\infty$
$\lim_{x \to \infty} x^3 - x^2 = \infty$
$\lim_{x \to \infty} (x + 2)^2 - (x^2 + 4x) = 4$

In each case, the expression we're taking the limit of is the indeterminate form $[\infty - \infty]$, but all three limits can be determined and all are different.

Yes. I have forgotten that $\infty - \infty$ is one of the indeterminate forms ...
Thank you for your reply.

Here's another way to solve the problem. Multiplying by a conjugate works in instances like this. Because $(\sqrt x - \sqrt 3)(\sqrt x + \sqrt 3) = x-3$, we can end up with a factor of $\sqrt{3-x}$ into every term so they will cancel out:
\begin{align*}\
\lim_{x \rightarrow 3^-} \frac { \sqrt {9-x^2} } {\sqrt x+\sqrt{3-x}-\sqrt3 }
&= \lim_{x \rightarrow 3^-} \frac { \sqrt {(3-x)(3+x)} } {\sqrt{3-x}-(\sqrt3-\sqrt x) } \cdot \frac{\sqrt{3} + \sqrt{x}}{\sqrt{3}+\sqrt{x}} \\
&= \lim_{x \rightarrow 3^-} \frac {(\sqrt{3} + \sqrt{x}) \sqrt {(3-x)(3+x)}}{(\sqrt{3} + \sqrt{x})\sqrt{3-x} -(3-x)} \\
&= \lim_{x \rightarrow 3^-} \frac {(\sqrt{3} + \sqrt{x}) \sqrt {3+x}}{\sqrt{3} + \sqrt{x} - \sqrt{3-x}}
\end{align*} Now there's no indeterminate form when you set $x=3$ so no need to use the Hospital rule to evaluate the limit. (Also, if you're allowed to use the Hospital rule, why not just use it right at the start?)

Thank you! I was trying to solve this without Hospital rule.