Question

find the limit of lim 4x^2-3 over or divide x-7 where x->2/3(two thirds)

Answer 1

replace x by 2/3 and see what you get. since you do not get 0/0that is your answer.

Answer 2

if the numerator is 0, and the denominator is not, then there is no limit

Answer 3

but it isn't so just plug in and see what you get

Answer 4

? If the numerator is 0 there's no answer? You mean, the denominator is 0?

If the numerator is 0, the limit is 0.

Answer 5

oh i think you should try factoring and i don't know the next step so can you help me..

Answer 6

yeah of course

Answer 7

my mistake.

Answer 8

@moniqu don't factor anything

Answer 9

replace x by
\[\frac{2}{3}\] and see what number you get. that is all

Answer 10

i get
\[\frac{11}{57}\]

Answer 11

it should be these one:\[(4\chi+3)(4\chi-3) \over \chi-7\]

Answer 12

ok lets go slow

Answer 13

you want
\[\lim_{x\rightarrow \frac{2}{3}}\frac{4x^2-3}{x-7}\] yes?

Answer 14

so 11 over 57 is correct?

Answer 15

this is a rational function and you are looking for the limit at a certain number. now a rational function is continuous everywhere on its domain. the domain of this thing is all numbers except 7, so if you want the limit anywhere else, just plug the number in and see what you get

Answer 16

i got 11/57 but you should check it. i just computed
\[\frac{4(\frac{2}{3})^2-3}{\frac{2}{3}-7}\]

Answer 17

same... but can I ask on what kind of given should I use factoring.?