Limit 1/x+1 as x approaches 0 find limit using delta epsilon definition

need help please

Assuming that we want to show that lim(x→0) 1/(x+1) = 1:

Given ε > 0, we want to find δ > 0 such that
0 < |x - 0| < δ ==> |1/(x+1) - 1| < ε.

To show this, note that
|1/(x+1) - 1|
= |x/(x+1)|
= |x| / (x+1), assuming that |x| < 1 (to remove absolute values)

Let's further assume that |x| < 1/2 to keep x completely away from -1 (and avoid division by 0 issues). So, |1/(x+1) - 1| = |x|/(x+1) < |x| / (-1/2 + 1) = 2|x|.
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Given ε > 0, let δ = min {1/2, ε/2}.
Then, 0 < |x - 0| < δ ==> |1/(x+1) - 1| < 2|x| < 2(ε/2) = ε.

Good luck!