can you help a b/rother out?
Can you help a b/rother out?
Substitute 2 for x and solve
DNE
Been a while since I took cal 1, but try using L'hospitals rule, that should do the trick.
22
0/0
big bang
factorise the upper exuation, assume 2 is a zero piont
4
Factors into (x-2)(x-3)/(x+2)(x-2)
Cancel (x-2)'s out
(x-3)/(x+2)
Substitute 2 in for x
-1/4?
And you can use some rule to slip the lower equation to (x-2)(x+2)
Factor the top and bottom if possible, plug and chug
wolframalpha.com
this
Thank me later OP.
5x/2?
Factorizing both, denominator and numerator, should give you:
(x - 3)(x - 2) / (x + 2) (x - 2) you can do it now
This
not how it's intended faggot. You don't just input values without having a basic understanding of what limits are.
Since this is a limit of the form 0/0, we use L'Hospital's rule, and get
f(x)=x^2-5x+6, f'(x)=2x-5.
g(x)=x^2-4, g'(x)=2x
If we follow the rule, it follows that
lim(f(x)/g(x)) as x->2 is equal to lim(f'(x)/g'(x)) as x->2, this gives
lim((x^2-5x+6)/(x^2-4) as x approaches 2 is equal to lim((2x-5)/(2x))=lim((2*2-5)/(2*2))=-1/4
So the result is -1/4
This
Calculus is fuuuun
Do what this guy says.
that's how limits work faggot. if you graph that expression against x you can clearly see that the expression approaches that value as x approaches 2
doesn't work, 0 denominator
doesn't work because you would have to apply it twice on the denomitator but the numerator doesn't go to 0 the second time.
instead you factorize as follows.
numerator: x^2 - 5x + 6 = (x - 2)(x - 3)
denominator: x^2 - 4 = (x - 2)(x + 2)
then cross out the (x - 2).
then shoe on head and sharpy in pooper.
this is a good way to test your answer, however, unless something funny is going on. But for most practical cases, it is fine.
I got this, Sup Forumsro, the answer is 3
calculus is cancer
wait, my bad, l'hôpital works just fine.
Factorising is easier though.
why would you have to do it twice on the denominator?
this guy did the whole thing for you
and it's the 'easiest' way of doing it, by a lot.
you'll eventually get questions where they cant be factorised.
Of course, that's the point of having a rule like L'Hospitals rule. To define limits of the form 0/0 or infinity/infinity.
yeah i know, realised it just after posting...
...
Just plug it into symbolab and it gives you step-by-step
Wolfram makes you pay for the step-by-step feature
symbolab.com