- #1

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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

- Thread starter evry190
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- #1

- 13

- 0

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

- #2

Office_Shredder

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You get to choose what d_{2} is

- #3

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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.

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