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you can just solve this limit by simplifying. But... To apply L'H rule just take the derivative of the top and bottom of the fraction and then simplify and try to sub the number in, if you still get 0/0 or infinity/infinity, apply L'H rule again by taking the derivative of the top and bottom again. Keep doing this until you can solve the limit.
I have to use L'H rule how can I take the derivative of the top and bottom? This is more complicated than we have learned in class.
You wont use L'H rule with this problem. because (x/2)(x-2)/1 = (x^(2)-2/2) Lim (x^(2)-2)/2 = 2/2 = 1 x->2
If you were using L'H in class then you would know it. You probably wont see L'H rule until calculus 1
if you are not in it if you are just watch a tutorial on youtube there are tons of videos explaining L'H rule
hurry up and reply i have a midterm in 20 minutes I have to get to
I am in calculus I. I know how to use L'H rule, I dont know how to take the derivative of this particular equation?
I made a mistake in the answer btw, I dont see how you can even apply L'H rule it is not a linear equation taht gives 0 and infinity and it is not a fraction taht gives 0/infinity
Solve this using simplification https://www.wolframalpha.com/input/?i=lim+x-%3E2+%28x%2F2%29%2F%281%2F%28x-2%29%29
Its ok if you cant help, nbd. It wouldnt surprise me if my teacher did something wrong he is a joke. Thanks!
yeah this problem doesnt require L'H rule, ask in chat to confirm but im 99.9% sure
click show steps on wolfram alpha it uses simplification as well
when I say differentiate the top and the bottom of the fraction I meant treat the denominator and neumorator as linear equations and take the derivative of them I recommend these videos this guy explain calculus concepts really well. It also helps if you are like me and day dream a lot during lectures. https://www.youtube.com/watch?v=PdSzruR5OeE www.youtube.com/watch?v=BiVOC3WocXs