Limit of function with rational

  • Thread starter iloveannaw
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  • #1
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Homework Statement


the title says it all

[tex]x \rightarrow 4[/tex]

for [tex]f\left(x\right) = \frac{\sqrt{1+2x}-3}{\sqrt{x}-2}[/tex]

I have multiplied both top and bottom by conjugate, [tex]\sqrt{x}+2[/tex]:

[tex]f\left(x\right) = \frac{\sqrt{x(1+2x)}+2\sqrt{1+2x} -3\sqrt(x)-6}{x-4}[/tex]


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

Answers and Replies

  • #2
Char. Limit
Gold Member
1,204
14

Homework Statement


the title says it all

[tex]x \rightarrow 4[/tex]

for [tex]f\left(x\right) = \frac{\sqrt{1+2x}-3}{\sqrt{x}-2}[/tex]

I have multiplied both top and bottom by conjugate, [tex]\sqrt{x}+2[/tex]:

[tex]f\left(x\right) = \frac{\sqrt{x(1+2x)}+2\sqrt{1+2x} -3\sqrt(x)-6}{x-4}[/tex]


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.
 
  • #3
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thankyou!!!
 
  • #4
Char. Limit
Gold Member
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No problem. Incidentally, since I didn't carry it all the way out myself, did it work?
 
  • #5
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yes, that's why I'm so pleased - it just needed a bit of factoring after doing it your way. cheers
 
  • #6
Char. Limit
Gold Member
1,204
14
Awesome!

Have a great day.
 

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