2 questions :) (1 Viewer)

Snowe

Just wandering around
Joined
Nov 11, 2004
Messages
37
Location
Sydney
Gender
Male
HSC
N/A
Hey guys have 2 questions :)

1) How do I solve (x+4)^4= 0?


Never seen this question before o.0



2) How do I find the limit of lim (x^2+3)/ (x+4) ?
x-> infinity


I know if it was a no. you would just substitute that no. inside because if you use a no. very close to that no. it would approach the no. being substituted but with the infinity its confusing o.0;
 

Xayma

Lacking creativity
Joined
Sep 6, 2003
Messages
5,953
Gender
Undisclosed
HSC
N/A
1)
(x+4)<sup>4</sup>=0
&there4;
(x+4)=<sup>4</sup>&radic;0
=0
&there4;
x=-4

2) lim (x--> &infin; ) (x<sup>2</sup>+3)/(x+4)
=lim (x--> &infin; ) (x<sup>2</sup>-16+19)/(x+4)
=lim (x--> &infin; ) (x<sup>2</sup>-16)/(x+4) +19/(x+4)
=lim (x--> &infin; ) (x+4)(x-4)/(x+4)+19/(x+4)
=lim (x--> &infin; ) (x-4)+19/(x+4)
as x--> &infin; 19/(x+4) --> 0
&there4;

lim (x--> &infin; ) (x^2+3)/(x+4)=x-4
which approaches &infin; as x--> &infin; the graph y=(x<sup>2</suP>+3)/(x+4) will approach y=x-4 as x--> &infin;
 
Last edited:

Estel

Tutor
Joined
Nov 12, 2003
Messages
1,261
Gender
Male
HSC
2005
by inspection the expression approaches infinity anyhow as the power of the numerator > power of the denominator.
 

Snowe

Just wandering around
Joined
Nov 11, 2004
Messages
37
Location
Sydney
Gender
Male
HSC
N/A
Hey man thanks heaps for the working out :D
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top