Proving the Limit of f(ax) with Delta Epsilon

[evry190]

hey if lim (x-->0) f(x) = L

where 0 < |x| < d1 implies |f(x) - L | < e

how do i prove lim (x --> 0) f(ax) = L?

i know

0 < |ax| < |a|d1

d2 = |a|d1

but the textbook says d2 = d1/|a|

help you guyssssssssssssssssssssssssssssssss

 

[Office_Shredder]

You get to choose what d₂ is

 

[Quantumpencil]

You know that the function f is continuous, so you know that given epsilon > 0, there exists an delta > 0 satisfying
d(x)<delta -> d(f(x),l))<epsilon. Now your problem asks you to show that, given an epsilon > 0, you can find a delta such that d(x)<delta -> d(f(ax),l)<epsilon.

The point is that, you can also choose such a delta for ax. the continuity of the f immediately tells you that you can choose such a delta. What if f were the function f(x) = x^3? consider a neighborhood of f(x)=l=0, (-8,8). Then if x is between -2,2, we know that this relation holds. However, f(ax), a = 5, gives us a drastically stretched function. (-8, 8) is stretched to (-40, 40). if x is between -2/5, 2/5 = original delta/a, then d (f(ax), l) < d(f(x),l)<epsilon, because d(x) is now less than the first delta (2) in this case (because f(ax), x<delta over a implies that x from our second delta neighborhood, times a, is in our first delta neighborhood Adapt, and you'll be done.