Question

How would I go about evaluating the limit as x approaches -1 of f(x) = (x^2 + 4x + 3) / (x + 1)? Obviously just putting in -1 doesn't work, and I'm not allowed to simply use a calculator to test values close to -1. So I'm unsure how else I'm supposed to go about this.
Also, I'm not even sure why there is even a function value at x=-1.
I tried drawing it in a graphing program and it's apparently actually a linear equation -- but how would I even know this? Can I simplify it somehow or something? Thanks.

Answer 1

\[\lim_{x \rightarrow -1}(x^2+4x+3)/(x+1)==>\]\[\lim_{x+1 \rightarrow 0}(x^2+4x+3)/(x+1)==>\]\[\lim_{t \rightarrow 0}((t-1)^2+4(t-1)+4)/t\]

Answer 2

\[\lim_{t \rightarrow 0}(t+1)^2/t\]

Answer 3

\[\infty\]

Answer 4

Sorry, but could you possibly elucidate what happened during and after the second step? What's the purpose of the x+1->0 and why does it suddenly change to t and... I'm perfectly confused. :S

Answer 5

i use t to express (x+1) and it is just experience. every time i meet this kind of questions,i always solve it like this.

Answer 6

t=x+1
just for convience

Answer 7

OK. And why do you change it to x+1->0?

Answer 8

first, x-->-1
and, below the line is (x+1)
i think it can be reckoned as a symbol.

Answer 9

do you understand it ?

Answer 10

But isn't the limit simply 2? I just realized I can probably simply use L'Hopital's Rule, meaning \[\lim_{x \rightarrow -1} (x^{2}+4x+3) / (x+1) = \lim_{x \rightarrow -1} 2x+4 = 2?\]

Answer 11

\[
\frac{x^2+4
x+3}{x+1}=\frac{(x+1)
(x+3)}{x+1}=x+3
\]

Answer 12

Limit is -1+ 3=2