You have to show :
hi, the only instruction is to show the function is continuous. i know the epsilon delta definition, but when that is required, usually it is explicitly mentioned. so, i am thinking i need to use a different idea. the function is defined as: arcsin(x)/x when x=/=0, and 1 when x=0. so, i am thinking about what method to use. should i show that when arcsin(x)/x approaches 0, the limit is 1? is this a sufficient condition? should i show that arcsinx/x is continuous elsewhere? or can i just state this (and possibly why?)
is l'opital essential here, i will be showing what with this? i understand the conditions are met for it, but, thn what? lopital shows differentiabiliy, which is the next question. so i need to show continuity here in a way which does not show differentiability.
To be differential at a point, the function needs to be continuous at that point AND reasonably smooth. To be smooth, that means that the gradients don't make sudden jumps. So the derivative ALSO needs to be continuous at that point.
i need to show that (arcsinx/x - 1 )/ x goes to 0 as x goes to 0. it should be l ' hopitalable but arcsin0/0 - 1 has to equal 0 for that to be so. in the original function, the function is defined as 1 at x=0, can i use this here to show the condition for l' hopital? even though its technically a new function? can i plug in that earlier definition?
L'Hopital's rule is used for the limit of a fraction - when you get or , you can take the derivative of the numerator and denominator and try again. So you have arcsin(x) for the numerator and x for the denominator. If you plug in the derivatives and set x=0, you should get the answer.
- Hollywood
absolutely, i know, but then if you use l'opital here. my issue is, can you use the fact that arcsinx/x was defined as 1 at 0, to say that the numerator is 0 at h=0. (thus l' hopital is valid).? or since its a 'new' function you can't do this. l'hopital
right, thats what i thought. but its the first line i wasnt sure about. whether i need to multiply through by h first, or if i could use the definition of the function to 'replace' the arcsinh/h with a 1 (since the function is defined as 1 at 0, therefore the function - 1 would be defined as 0 at 0). though., the net result is the same. thats actually what i did. to be honest. what you wrote above is exactly what i did. thanks for the confirmation.