Help whit his limit

[Guest]

[tex]\lim_{x \to 0} \frac{\sqrt{3x-2}+\sqrt[3]{x+6}-4}{x-2}[/tex]
Without L'hopital's Rule

[Guest]

[tex]\lim_{x \to 2} \frac{\sqrt{3x-2}+\sqrt[3]{x+6}-4}{x-2}[/tex]
Without L'hopital's Rule

[Guest]

That's pretty standard- rationalize the numerator! The "non-standard" part is that the numerator is rather complicated but we can "uncomplicate" it.

Noting that [tex]\sqrt{3x- 2}[/tex] goes to [tex]\sqrt{6- 2}= \sqrt{4}= 2[/tex] and [tex]\sqrt[3]{x+ 6}[/tex] goes to [tex]\sqrt[3]{2+ 6}= \sqrt[3]{8}= 2[/tex] I would write this as
[tex]\frac{\sqrt{3x-2}+ \sqrt[3]{x+ 6}- 4}{x- 2}[/tex][tex]= \frac{(\sqrt{3x-2}- 2)+ (\sqrt[3]{x+ 6}- 2)}{x- 2}[/tex]
[tex]= \frac{\sqrt{3x-2}- 2}{x- 2}+ \frac{\sqrt[3]{x+ 6}- 2}{x- 2}[/tex]
and do each fraction separately.

[tex]\frac{\sqrt{3x- 2}- 2}{x- 2}\frac{\sqrt{3x- 2}+ 2}{\sqrt{3x-2}+2}= \frac{3x- 2- 4}{(x- 2)(\sqrt{3x- 2}+ 2)}[/tex]
[tex]= \frac{3x- 6}{(x-2)(\sqrt{3x- 2}+ 2}= \frac{3(x- 2)}{(x- 2)(\sqrt{3x- 2}+ 2)}[/tex].

As long as x is not 2, we can cancel the "x- 2" in numerator and denominator and have [tex]\frac{3}{\sqrt{3x- 2}+ 2}[/tex]. Since that consists of continuous functions and the denominator now is not 0 at x= 2, we can take the limit, as x goes to 2, by setting x= 2: [tex]\frac{3}{\sqrt{3(2)- 2}+ 2}= \frac{3}{4}[/tex].

For [tex]\lim_{x\to 2}\frac{\sqrt[3]{x+ 6}- 2}{x- 2}[/tex] use the fact that [tex](x- a)(x^2+ ax+ a^2)= x^3+ ax^2+ a^2x- ax^2- a^2x- a^3= x^3- a^3[/tex] and multiply
[tex]\frac{\sqrt[3]{x+6}- 2}{x- 2}\frac{\sqrt[3]{x+ 6}^2+ 2\sqrt[3]{x+ 6}+ 4}{\sqrt[3]{x+ 6}^2+ 2\sqrt[3]{x+ 6}+ 4}[/tex]
[tex]= \frac{x+ 6- 8}{(x- 2)((x- 2)(\sqrt[3]{x+ 6}^2+ 2\sqrt[3]{x+ 6}+ 4)}= \frac{x- 2}{(x- 2)(\sqrt[3]{x+ 6}^2+ 2\sqrt[3]{x+ 6}+ 4)}[/tex]

As long as x is not 0, we can cancel the "x- 2" in numerator and denominator and have
[tex]\lim_{x\to 0}\frac{1}{\sqrt[3]{x+ 6}^2+ 2\sqrt[3]{x+ 6}+ 4)}[/tex]

Since the functions are continuous and the denominator does not go to 0, we can take the limit by setting x= 2:
[tex]\frac{1}{\sqrt[3]{2+ 6}^2+ 2\sqrt[3]{2+ 6}}[/tex][tex]= \frac{1}{2^2+ 2(2)+ 4}= \frac{1}{12}[/tex].

And, finally, add those two limits.