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What is the limit of` (x^2+3)/x^3` as x approaches infinite?

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ogidi | Student | Honors

Posted September 2, 2013 at 1:55 AM via web

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What is the limit of` (x^2+3)/x^3` as x approaches infinite?

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Mary Joy Ripalda | High School Teacher | (Level 3) Educator

Posted September 2, 2013 at 2:12 AM (Answer #1)

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`lim_(x->oo)` `(x^2+3)/x^3`

To determine the limit of a rational function when x approaches infinity, apply the property:

`lim_(x->oo) `   `1/x^n=0`

To do so, multiply the numerator and the denominator reciprocal of the term with highest exponent. And simplify. 

`=lim_(x->oo)` `(x^2+3)/x^3*(1/x^3)/(1/x^3) `

`=lim_(x->oo)` `(x^2/x^3+3/x^3)/(x^3/x^3)`

`=lim_(x->oo)` `(1/x+3/x^3)/1 `                          

`=lim_(x->oo)` `(1/x+3/x^3)`

`=lim_(x->oo)1/x+lim_(x->oo)3/x^3`

Now that it is in simplified form, apply the property above.

`=0+0`

`=0`

Hence, `lim_(x->oo)(x^2+3)/x^3=0` .          

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freemihai | College Teacher | (Level 1) Adjunct Educator

Posted September 3, 2013 at 5:52 AM (Answer #2)

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All polynomial limits we want to solve start with the simple verification: is or isn't an indeterminate form?

Please, start the verification by plugging oo in place of x, the same as we do when we want to calculate the value of polynomial:

`lim(x->oo) (x^2+x)/x^3 = (oo^2+oo)/(oo^3)`

There is no reason to say that `oo^3 = oo^2 = oo` .

Oops! It looks like the limit goes into an indeterminate form oo/oo. But it is not a problem when the numerator and denominator are differentiable functions, because, we can make use of l'Hospital's rule. Watch out!

What does this  l'Hospital's rule to help use?  Watch out the description!

L'Hospital's rule considers numerator f(x) and denominator g(x) and then mit makes the statement:

`lim(x->oo) f(x)/g(x) = lim(x->oo) (f'(x))/(g'(x)) =lim(x->oo) ` `(f''(x))/(g''(x))....`

Repeat the differentiation as many times as it should till the limit becomes determinate and it can be calculated.

Let's differentiate the numerator f(x) and denominator g(x):

`f'(x) = (x^2+x)' g'(x) = (x^3)'`

`f'(x) = 2x+1g'(x) = 3x^2`

Put the derivatives at their places and calculate the limit of derivatives:

`lim(x->oo) (2x+1)/(3x^2) = (2oo+1)/(3oo^2) = oo/oo`

Oops! Indeterminate form! Repeat l'Hospital's rule!

`f'(x) = 2x+1g(x) = 3x^2`

`f''(x) = 2 g''(x) = 6x`

Put the derivatives at their places and calculate the limit of derivatives:

`lim(x->oo) 2/6x = 1/(3oo) = 1/oo -> 0`

Notice division by oo of any constant gives 0.

The answer to the limit is `lim(x->oo) (x^2+x)/x^3`  = 0.

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