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Limit question with radicals

bnosam

New member
Sep 8, 2013
9
\(\displaystyle \lim_{x \to a} \frac{ x^2 - a^2}{\sqrt(x) - \sqrt(a)}\)

I've tried to solve this standard, but I either end up with 0 in the denominator, or I end up with 0/0.

Any hints on what to do with this next?

Thanks
 

chisigma

Well-known member
Feb 13, 2012
1,704
\(\displaystyle \lim_{x \to a} \frac{ x^2 - a^2}{\sqrt(x) - \sqrt(a)}\)

I've tried to solve this standard, but I either end up with 0 in the denominator, or I end up with 0/0.

Any hints on what to do with this next?

Thanks
Why don't multiply numerator and denominator by $\displaystyle \sqrt{x} + \sqrt{a}$?...

Kind regards

$\chi$ $\sigma$
 

bnosam

New member
Sep 8, 2013
9
Why don't multiply numerator and denominator by $\displaystyle \sqrt{x} + \sqrt{a}$?...

Kind regards

$\chi$ $\sigma$
I tried that, however I end up with x - a, at the bottom, which leads to 0 in the denominator.


\(\displaystyle \frac{x^2\sqrt(x) + x^2 \sqrt(a) - a^2\sqrt(x) - a^2\sqrt(a)}{x-a}\)
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
I suggest factorizing

\(\displaystyle (x^2-a^2)=(x-a)(x+a)=(\sqrt{x}-\sqrt{a})(\sqrt{x}+\sqrt{a})(x+a)\)
 

chisigma

Well-known member
Feb 13, 2012
1,704
I tried that, however I end up with x - a, at the bottom, which leads to 0 in the denominator.


\(\displaystyle \frac{x^2\sqrt(x) + x^2 \sqrt(a) - a^2\sqrt(x) - a^2\sqrt(a)}{x-a}\)
$\displaystyle \frac{x^{2} - a^{2}}{\sqrt{x}-\sqrt{a}}\ \frac{\sqrt{x} + \sqrt{a}}{\sqrt{x} + \sqrt{a}} = \frac{x^{2}-a^{2}}{x-a}\ (\sqrt{x}+\sqrt{a})= (x+a)\ (\sqrt{x}+\sqrt{a})$

Kind regards

$\chi$ $\sigma$
 

bnosam

New member
Sep 8, 2013
9
$\displaystyle \frac{x^{2} - a^{2}}{\sqrt{x}-\sqrt{a}}\ \frac{\sqrt{x} + \sqrt{a}}{\sqrt{x} + \sqrt{a}} = \frac{x^{2}-a^{2}}{x-a}\ (\sqrt{x}+\sqrt{a})= (x+a)\ (\sqrt{x}+\sqrt{a})$

Kind regards

$\chi$ $\sigma$
Ohh ok, I should have seen that.

\(\displaystyle x\sqrt(x) + x\sqrt(a) + a\sqrt(x) + a\sqrt(a)\)

\(\displaystyle = 4a\sqrt(a)\)

That seem right?
 

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
Ohh ok, I should have seen that.

\(\displaystyle x\sqrt(x) + x\sqrt(a) + a\sqrt(x) + a\sqrt(a)\)

\(\displaystyle = 4a\sqrt(a)\)

That seem right?
It should technically be \[\color{red}{\lim\limits_{x\to a}} x\sqrt{x}+x\sqrt{a}+a\sqrt{x}+a\sqrt{a} = 4a\sqrt{a}\]
since you're supposed to include the $\displaystyle\lim_{x\to a}$ part in each line of your work leading up to the substitution of $x=a$ at the end of the problem.

Otherwise, everything looks fine to me!
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
\(\displaystyle \lim_{x \to a} \frac{ x^2 - a^2}{\sqrt(x) - \sqrt(a)}\)

I've tried to solve this standard, but I either end up with 0 in the denominator, or I end up with 0/0.

Any hints on what to do with this next?

Thanks
Just a $\LaTeX$ tip:

Use the code \sqrt{x} instead of \sqrt(x) and the argument will be put under the radical, to get $\sqrt{x}$ instead of $\sqrt(x)$.