MHB How Do You Solve Limit Problems with Radicals and Zero in the Denominator?

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To solve the limit problem involving radicals and a zero denominator, one effective approach is to multiply both the numerator and denominator by the conjugate of the denominator, which is \(\sqrt{x} + \sqrt{a}\). This manipulation simplifies the expression and helps eliminate the zero in the denominator. After applying this technique, the limit can be evaluated by factoring the numerator, \(x^2 - a^2\), into \((x - a)(x + a)\). Ultimately, substituting \(x = a\) yields a final result of \(4a\sqrt{a}\). Proper notation in LaTeX is also recommended for clarity in mathematical expressions.
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$$\lim_{x \to a} \frac{ x^2 - a^2}{\sqrt(x) - \sqrt(a)}$$

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
 
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bnosam said:
$$\lim_{x \to a} \frac{ x^2 - a^2}{\sqrt(x) - \sqrt(a)}$$

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

$\chi$ $\sigma$
 
chisigma said:
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.$$\frac{x^2\sqrt(x) + x^2 \sqrt(a) - a^2\sqrt(x) - a^2\sqrt(a)}{x-a}$$
 
I suggest factorizing

$$(x^2-a^2)=(x-a)(x+a)=(\sqrt{x}-\sqrt{a})(\sqrt{x}+\sqrt{a})(x+a)$$
 
bnosam said:
I tried that, however I end up with x - a, at the bottom, which leads to 0 in the denominator.$$\frac{x^2\sqrt(x) + x^2 \sqrt(a) - a^2\sqrt(x) - a^2\sqrt(a)}{x-a}$$

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

Kind regards

$\chi$ $\sigma$
 
chisigma said:
$\displaystyle \frac{x^{2} - a^{2}}{\sqrt{x}-\sqrt{a}}\ \frac{\sqrt{x} + \sqrt{a}}{\sqrt{x} + \sqrt{a}} = \frac{x^{2}-a^{2}}{x-a}\ (\sqrt{x}+\sqrt{a})= (x+a)\ (\sqrt{x}+\sqrt{a})$

Kind regards

$\chi$ $\sigma$

Ohh ok, I should have seen that.

$$x\sqrt(x) + x\sqrt(a) + a\sqrt(x) + a\sqrt(a)$$

$$= 4a\sqrt(a)$$

That seem right?
 
bnosam said:
Ohh ok, I should have seen that.

$$x\sqrt(x) + x\sqrt(a) + a\sqrt(x) + a\sqrt(a)$$

$$= 4a\sqrt(a)$$

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!
 
bnosam said:
$$\lim_{x \to a} \frac{ x^2 - a^2}{\sqrt(x) - \sqrt(a)}$$

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 $\sqrt(x)$.
 

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