Register to reply

How about this limit?

by ladyrae
Tags: limit
Share this thread:
ladyrae
#1
Jun4-04, 06:26 PM
P: 32
Thanks...

Evaluate the limit

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

I simplified to (2x^2-3)/(2x+1) = -1/-1 = 1

Is this correct?
Phys.Org News Partner Science news on Phys.org
Study links polar vortex chills to melting sea ice
Lab unveil new nano-sized synthetic scaffolding technique
Cool calculations for cold atoms: New theory of universal three-body encounters
Zorodius
#2
Jun4-04, 06:42 PM
P: 184
I believe you want to factor the top, cancel out the hole in the graph, and then evaluate by substitution.

[tex]\lim_{x\rightarrow -1} \frac {2x^2-x-3}{x+1}[/tex]

[tex]= \lim_{x\rightarrow -1} \frac {(2x-3)(x+1)}{x+1}[/tex]

[tex]= \lim_{x\rightarrow -1} 2x-3[/tex]

= -5.
Caldus
#3
Jun4-04, 08:47 PM
P: 106
You can use L'Hospital's rule:

[tex]\lim_{x\rightarrow -1} \frac {2x^2-x-3}{x+1}[/tex]

If you plug in, you get 0/0, so:

[tex]\lim_{x\rightarrow -1} \frac {2x^2-x-3}{x+1}[/tex]
= [tex]\lim_{x\rightarrow -1} \frac {4x-1}{1}[/tex]

Plug in and you get -5.


Register to reply

Related Discussions
Limit of sequence equal to limit of function Calculus 1
Why is the limit 2025? a simple plug and chug limit! gone wrong! Calculus & Beyond Homework 3
Iterated limit? double limit? Introductory Physics Homework 4
How to prove there is no limit of cos1/x using theorm of limit Calculus 9