Limit of function with rational

[iloveannaw]

Homework Statement

the title says it all

$$x \rightarrow 4$$

for $$f\left(x\right) = \frac{\sqrt{1+2x}-3}{\sqrt{x}-2}$$

I have multiplied both top and bottom by conjugate, $$\sqrt{x}+2$$:

$$f\left(x\right) = \frac{\sqrt{x(1+2x)}+2\sqrt{1+2x} -3\sqrt(x)-6}{x-4}$$but don't know how to take this further. Dividing both numerator and denominator by x doesn't help.

 

[Char. Limit]

Homework Statement

the title says it all

$$x \rightarrow 4$$

for $$f\left(x\right) = \frac{\sqrt{1+2x}-3}{\sqrt{x}-2}$$

I have multiplied both top and bottom by conjugate, $$\sqrt{x}+2$$:

$$f\left(x\right) = \frac{\sqrt{x(1+2x)}+2\sqrt{1+2x} -3\sqrt(x)-6}{x-4}$$


but don't know how to take this further. Dividing both numerator and denominator by x doesn't help.

Try rationalizing the numerator first, then the denominator. That's what I would do.

 

[iloveannaw]

thankyou!

 

[Char. Limit]

No problem. Incidentally, since I didn't carry it all the way out myself, did it work?

 

[iloveannaw]

yes, that's why I'm so pleased - it just needed a bit of factoring after doing it your way. cheers

 

[Char. Limit]

Awesome!

Have a great day.