I'm currently studying for my Analysis exam, but can't seem to get the proof for the following limit.

Let $f : ℝ→ ℝ$ be defined by: $$f(x) = x^2 − |x|, x ∈ ℝ$$ Using the definition of limit of a real function at a point, show that : $$\lim_{x \to c} f(x) = c^2 - |c|,c∈ ℝ$$ I have been able to use the delta-epsilon proofs in other examples but am struggling with this one.


Choose $c\in \mathbb{R}$ and choose $\varepsilon >0$. Then $$|f(x)-f(c)|=|x^2-c^2+|c|-|x||\leq |x^2-c^2|+||c|-|x||\leq |x-c||x+c|+|x-c|$$ for all $x$. Here we used the triangle inequality and the reverse triangle inequality. If $|x-c|< \delta$, we see that $$|f(x)-f(c)|< \delta(|x+c|-1).$$

I'll leave the rest up to you. Use the $|x-c|<\delta$ to find an upperbound of $|x+c|$ in terms of $\delta$. Then choose $\delta$ appropriately to make $|f(x)-f(c)|<\varepsilon$.

  • $\begingroup$ That's great, thank you! $\endgroup$ – Cillian328 May 14 '16 at 11:44

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