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

Asked on by ogidi

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freemihai's profile pic

freemihai | College Teacher | (Level 2) Adjunct Educator

Posted on

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.

lemjay's profile pic

lemjay | High School Teacher | (Level 3) Senior Educator

Posted on

`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)`


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



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

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