Question

Just the ones without answers plz

3x Consider the hyperbola f(x) = The numerator is dominated by 3x and the numerator is dominated by x, so we can easily convince ourselves that the limit of this function as x goes to infinity is L = lim f(x) =3 Now we prove this using the formal definition of a limit. Given any e 0 assume x> M.Since M can be large we also assume that M> 2 So: Step 1. Using your value for L above, simplify If(x) - L in terms of x f (x)- L 6/(x-2) Note that, for x > 2, the expression you've found strictly decreases as x increases. Syntax help: the Maple syntax for lxl is abs (x) 2. Use the assumption that x > M (and M > 2) to obtain an upper bound on f(x) - L in terms of M f(x)- L< 3, Choose M so that your answer in step two equals c: Syntax help: recall the syntax for e epsilon. Let's repeat the above argument for g(x) = _2x x-3 A formal proof f the statement mg(x) = L = 2 is as follows. Given any e > 0 ifx > M> 3 then g(x)- LE provided Note: the Maple syntax for xl is abs (x), and the syntax for e is epsilon

Answer 1

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