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

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SUMMARY

The discussion focuses on solving the limit problem $$\lim_{x \to a} \frac{x^2 - a^2}{\sqrt{x} - \sqrt{a}}$$, which initially results in an indeterminate form of 0/0. Participants suggest multiplying the numerator and denominator by $$\sqrt{x} + \sqrt{a}$$ to simplify the expression. This leads to the factorization of the numerator as $$(x-a)(x+a)$$, allowing for cancellation and the evaluation of the limit to yield $$4a\sqrt{a}$$. Additionally, a LaTeX formatting tip is provided for proper representation of square roots.

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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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