Question

how to find this limit: lim (xˆ2-1)/(x-1) when x goes to 1?

Answer 1

\[\lim_{x \rightarrow 1}(x ^{2}-1)/x-1\]

Answer 2

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

Answer 3

remember there is a hole at F(1)

Answer 4

remember x never equal to 1 but all ways in deleted neighbourhood 1

Answer 5

2 anavega
yeah, we have a hole, but we also have a problem 0/0 =)
but we can do the dividing 'cause we have lim
when we work with lim we have extra privileges

Answer 6

remember the concept of lim works only on deleted neighbour of a
\[\left( a-\xi,a+\xi \right)\ \left\{ a \right\}\]

Answer 7

a is not there in the set

Answer 8

so cannot get 0/o form

Answer 9

okay, you know about the Hopital's rule?
http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule

it say
\[\lim_{x \rightarrow c} f(x)/g(x) = [0/0] = \lim_{x \rightarrow c} f(x) \prime/g(x) \prime\]

if you do that
we should take
\[\lim_{x \rightarrow 1} (x ^{2}-1)/(x-1) =\lim_{x \rightarrow 1} 2x = 2\]
isn't it?

Answer 10

it is not =0/=0 cheak
it is\[\rightarrow0/\rightarrow0\]
=0/=0 is undefined
but \[\rightarrow0/\rightarrow0\]
is indeterminate form

Answer 11

In calculus, Hopital's rule p (also called Bernoulli's rule) uses derivatives to help evaluate limits involving indeterminate forms.

Answer 12

ha then it is \[\epsilon/\epsilon\]
epsilon\[\epsilon \rightarrow0\]

Answer 13

so we cancel the \[\epsilon\]

Answer 14

you mean that
\[\lim_{\epsilon \rightarrow 0} \epsilon/\epsilon = 1\]

Answer 15

yes

Answer 16

yeah, you are right, 'cause in this case we have indeterminate form like you say previously
\[\lim_{x \rightarrow 0} x/x = [0/0] use Hopital's rule = \lim_{x \rightarrow 0} 1/1 = 1\]

Answer 17

you can use the l'Hôpital's rule。 lim (xˆ2-1)/(x-1) it is the form of 0/0, because if we make x=1, so x^2-1=0, x-1=0, so it is the form of 0/0
so, at that time we can use l'Hôpital's rule to differentiate both denominator and numerator. so lim (xˆ2-1)/(x-1) =d(x^2-1)/d(x-1)=2x=2

Answer 18

L'hopital's Rule is okay, but here there is an even easier answer
x^2-1=(x+1)(x-1) by factoring so:
(x^2-1)/(x-1)=x+1
Then, when x goes to 1 it is clear that f(x) goes to 2.