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Solve algebraically using L'Hopital's Rule and please explain:...

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user1928785 | eNotes Newbie

Posted April 26, 2013 at 3:36 AM via web

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Solve algebraically using L'Hopital's Rule and please explain:

`lim_(x->oo)sqrt(4x^2+1)/(x+1)`

 

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oldnick | Valedictorian

Posted April 26, 2013 at 2:54 PM (Answer #2)

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First we se:

`lim_(x->oo)(sqrt(4x^2+1))/x+1=lim_(x->oo) (sqrt(4+ 1/x^2))/(1+1/x)=2`

Appling De l'Hopital rule we get:

`lim_(x->oo)(sqrt(4x^2+1))/(x+1)=lim_(x->oo) 4x/(sqrt(4x^2+1))`

thus:

`lim_(x->oo) (sqrt(4x^2+1))/(x+1)= 4 lim_(x->oo) (x/x)/sqrt(4+1/x^2)` `=4 xx 1/2=2`

that  the outome is equal despite the rule we've used.

 

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oldnick | Valedictorian

Posted April 26, 2013 at 2:58 PM (Answer #3)

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numerator and denominator with their derivatives. The result is:

=

This limit is 2 no  `oo/oo`  !

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted April 26, 2013 at 3:55 AM (Answer #1)

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The value of `lim_(x->oo)sqrt(4x^2+1)/(x+1)` has to be determined.

Substituting `x = oo` gives the indeterminate form `oo/oo` . This allows us to use l'Hopital's rule and substitute the numerator and denominator with their derivatives. The result is:

`lim_(x->oo)(8*x)/(2*sqrt(4x^2+1))`

= `lim_(x->oo)(4*x)/sqrt(4x^2+1)`

Substituting `x = oo` again gives the indeterminate form `oo/oo` . Use l'Hopital's rule:

`lim_(x->oo) 4/((4*x)/sqrt(4*x^2+1))`

= `lim_(x->oo) sqrt(4x^2+1)/x `

= `lim_(x->oo) sqrt((4x^2)/x^2+1/x^2)`

= `lim_(x->oo) sqrt(4+1/x^2)`

As `x->oo` , `1/x->0`

= `sqrt 4`

= 2

The limit `lim_(x->oo)sqrt(4x^2+1)/(x+1) = 2`

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