Proving \lim_{x \to a} f(x) = \lim_{h \to 0} f(a+h): Analysis & Feedback

[jgens]

So, I'm trying to prove that \lim_{x \to a} f(x) = \lim_{h \to 0} f(a+h) and I'm wondering if this "proof" is correct:

Suppose that \lim_{x \to a} f(x) = l and \lim_{h \to 0} f(a+h) = m with l \neq m. Then, for any \varepsilon > 0 there must be some \delta > 0 such that 0 < |x-a| < \delta implies that |f(x) - l| < \varepsilon and 0< |h| < \delta that |f(a+h) - m| < \varepsilon.

Clearly a - \delta < x < a + \delta and a - \delta < a + h < a + \delta, so there is a number x_1 such that |f(x_1) - l| < \varepsilon and |f(x_1) - m| < \varepsilon.

Choosing \varepsilon = |l-m|/2, it follows that |l-m| = |l - f(x_1) + f(x_1) -m| \leq |f(x_1) - l| + |f(x_1) - m| < |l-m|/2 + |l-m|/2 < |l-m|, which is a contradiction. Therefore, l=m.

Does this work alright? Or is there another way that I should approach this problem? I appreciate any feedback! Thanks!

 

[snipez90]

Your approach seems to work, but I think it would have been much easier to write out what one of the limits means and then later set x = a + h. This approach also supports the intuitive idea behind this equality. There is also a way to prove this without resorting to the definition of the limit via some manipulations.

 

[losiu99]

Looks ok. It is possible to do it more directly. Suppose \lim_{x\to a}f(x)=L. Fix \epsilon > 0. For some \delta-neighbourhood of a, |f(x)-L|<\epsilon. Hence, if |h|<\delta, |f(a+h)-L|<\epsilon holds, and therefore \lim_{h\to 0}f(a+h)=L. It's essentially the same, though.

Edit: I'm too slow
Edit2: Wow, 3 minutes...way too slow