Calculus: Limits

[KidInkFan]

Let f(x)=9

x6−15

x3+8 and g(x)=c

xn−2

x3+15 with c

=0. Thenlimx

f(x)/g(x) = infinity



I thought that the degree in the denominator had to be < then the degree in the numerator in order for the limit to be infinity. As well I thought 'c' would
be denoted as a value less than 9. So c < 9 and n < 6 but im still not getting it right


Thanks for anyone's help towards this problem!

 

[Deleted member 4993]

Let f(x)=9

x6−15

x3+8 and g(x)=c

xn−2

x3+15 with c

=0. Then limx

f(x)/g(x) = infinity



I thought that the degree in the denominator had to be < then the degree in the numerator in order for the limit to be infinity. correct

As well I thought 'c' would be denoted as a value less than 9. incorrect

It is already given to you → \(\displaystyle c \ne 0\)

So c < 9 and n < 6 but im still not getting it right


Thanks for anyone's help towards this problem!

.

 

[HallsofIvy]

Let f(x)=9

x6−15

x3+8 and g(x)=c

xn−2

x3+15 with c

=0. Thenlimx

f(x)/g(x) = infinity
First, at least us "^" to indicate powers You have (9x^6- 15x^3+ 8)/(cx^n- 2x^3+ 15).



I thought that the degree in the denominator had to be < then the degree in the numerator in order for the limit to be infinity. As well I thought 'c' would
be denoted as a value less than 9. So c < 9 and n < 6 but im still not getting it right

Yes, in order that the limit be infinity, the numerator must have greater power so you must have n< 6. However, once that is true, the value of c, other than the fact that it is non-zero, is irrelevant.

Thanks for anyone's help towards this problem!

 

[KidInkFan]

Thanks I figured it out!! The 'c' had to be a value > 0 otherwise it would be a limit approaching negative infinity if it were < 0!

 

[HallsofIvy]

Yes, if you are distinguishing between \(\displaystyle \infty\) and \(\displaystyle -\infty\), that is true. That distinction is not always made by just saying "infinity".