Indeterminate Forms for L'Hopital Rule: Calculus?

Question

Is 0-0 and inf + inf indeterminate forms
please explain

Thank You

Answer 1

Solution:

You apply L'hospital rule when their is indeterminacy of 0/0 and infinity/ infinity . Thats all.

0-0 is nothing but 0 and inf + inf is infinity so they are not indeterminable .

HOWEVER 0* infinity ,0^ infinity is indeterminable but still L'hospital is not applicable. To

apply l;hos rule you must first convert it into above two indeterminacy.

Answer 2

no,

0-0=0 and inf + inf = inf. This is not indeterminate because we can, in a sense, "determine" the result.



when we look at division tho,

0/0 is undefined and when taking limits that results in this behavior we cannot determine what the ratio itself converges to directly. same with inf/inf. In essence, when we get something in an indeterminate form, we then want to analyze the rate at which the numerator and the denominator changes as "x" changes. for example, take 2x/x. it may seem trivial at first glance but observe how 2x increases twice as fast as x for all x as x increases. This time you get inf/inf which is indeterminate but L'Hopital's rule allows you to get this to converge to 2.

Answer 3

For l'Hopital's rule, the indeterminate forms are 0/0 and inf/inf. Neither 0/0 nor inf/inf yields a defined result. Therefore, they are indeterminate.



0-0 and inf+inf do not qualify for use of l'Hopital's rule since 0-0 evaluates to 0 and inf+inf evaluates to inf.

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Answer 4

I'd think 0 - 0 and ∞ + ∞ wouldn't be indeterminate forms because they both either approach 0 or ∞. The only indeterminate forms I know of are 0^0, 0^∞, ∞^0, which must use logs to simplify to 0/0 or ∞/∞ in order to use L'Hopital's Rule.

Answer 5

No.



If any limit results in 0 - 0, then the limit itself is zero.



If any limit results in ∞ + ∞, then the limit itself is ∞.

Answer 6

ln(2x) - ln(x+a million) = ln (2x/(x+a million)) the two ln and (2x/(x+a million) are non-supply up and tender (do no longer keep in mind how lots of this is honestly needed) so we are able to equate lim ln f(x) = ln lim f(x) go away ln out for now, l'well being center says that lim (f(x)/g(x)) would be lim (df/dx)/(dg/dx) each so often alongside with infinity divided by using infinity. so lim (2x/(x+a million)) = lim ( ((2x)/dx) / ((x+a million)/dx) ) = lim(2/a million) = 2 lim ln (2x/(x+a million)) =ln lim (2x/(x+a million)) = ln 2