# Find an open interval about xnaught on which the inequality lf(x)-Ll < epsilon holds.then give a value for delta > 0 such that for all x satisfying 0 < lx-xnaughtl < delta the...

Find an open interval about xnaught on which the inequality lf(x)-Ll < epsilon holds.

then give a value for delta > 0 such that for all x satisfying 0 < lx-xnaughtl < delta the inequality lf(x)-Ll < epsilon holds.

f(x) = mx+b, m>0, L=m+b, xnaught = 1, epsilon =0.05

A simple explination of this problem would be fantastic.

### 1 Answer | Add Yours

Given F(x)=mx+b, L=m+b, `epsilon=.05` , `x_0=1` ; Find `delta` so that

`0<|x-x_0|<delta -> |f(x)-L|<epsilon`

We are asked to find a delta neighborhood about 1 (`1-delta,1+delta` ), so that the values for the function within that neighborhood are within 0.05 of the limit, given as m+b.

Let `delta=.05/|m|` . Now `0<|x-1|<delta -> 0<|x-1|<.05/|m|` . Then `|m||x-1|<.05` .

Now `|m(x-1)|<.05` and `|mx-m|<.05` . Suppose we add and subtract b inside the absolute value -- this is adding zero so creates an equivalent expression thus `|mx+b-m-b|<.05` or `|mx+b-(m+b)|<.05` which is `|f(x)-L|<epsilon` .

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Some magic seemed to occur -- where did the let `delta=.05/|m|` come from? Just work the problem backwards; start with |f(x)-L|<`epsilon` and manipulate the left-hand side until you get `|x-x_0|` , the expression on the right-hand side will be the delta you are seeking. This is also how we knew to add/subtract b -- we needed it for the final expression.