The following is a few proofs of limits of functions. There are mistakes in the proofs. Can you find where the mistakes are? Can you fix them?
Problem 1. Use the formal definition to prove the limit: \[\lim_{x\to 5 -} \frac{x}{x-5} = - \infty\]
Solution(including mistakes). Let \(B > 0\) be given. We have to find \(\delta > 0\) such that for all \(x\) in the domain, \[ \frac{x}{x-5} < -B \quad \text{whenever} \quad -\delta < x-5 < 0 .\] Assuming that \(\delta \le 1,\) we have \(4 < x < 5\) which implies \[ \frac{x}{x-5} < \frac{5}{x-5}.\] In order to derive the desired inequality, we have to find \(\delta >0\) such that \[\frac{5}{x-5} < \frac{5}{-\delta} < -B;\] From this inequality, we see that \(\delta < 5/B\) is the other condition.
Thus \(\delta = \min \left\{ 1 ,\, 5/B \right\}\) works.
Problem 2. Use the formal definition to prove the limit: \[\lim_{x\to 1} \lfloor x \rfloor (x-1) =0\] where \(\lfloor x \rfloor\) is the greatest integer function.
Solution(including mistakes). If \(x\) is near \(1,\) \(\lfloor x \rfloor\) assumes only \(0\) or \(1;\) We have the following inequality: \[\left\lvert \lfloor x \rfloor (x-1) \right\rvert \le 1 \cdot \left \lvert x-1 \right \rvert.\] Now let \(\epsilon > 0\) be given. Take \(\delta = \epsilon .\) If \(0 < \lvert x-1 \rvert < \delta,\) we have \[\left\lvert \lfloor x \rfloor (x-1) -0 \right\rvert \le 1 \cdot \left \lvert x-1 \right \rvert < 1 \cdot \delta = \epsilon,\] which is the desired inequality.
For the first one, (x-5) < 0 and (0 <) x \frac{5}{x-5} in fact. (dividing by a negative number changes the direction of inequality). One has to use the following correct inequality:
\frac{x}{x-5} < \frac{4}{x-5}
For the second one, I'm not sure tbh.