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