Find $$\lim_{t\to 0}\frac{|t-2|}{t}$$ and $$\lim_{t\to\infty}\frac{|t-2|}{t}$$

Usually I would simply the top and bottom but I'm not sure what to do for absolute values.

Any help would be appreciated. Thanks!


1 Answer 1


HINT : Note that $$|t-2|=\begin{cases}t-2&\text{if $t-2\ge 0$}\\-(t-2)&\text{if $t-2\lt 0$}\end{cases}$$ Hence, we have $$\lim_{t\to 0}\frac{|t-2|}{t}=\lim_{t\to 0}\frac{-(t-2)}{t}$$ and

$$\lim_{t\to \infty}\frac{|t-2|}{t}=\lim_{t\to\infty}\frac{t-2}{t}.$$

  • $\begingroup$ Okay, thanks. How would I solve for g'(1) and g'(2)? I'm not sure how the prime works with limits? $\endgroup$
    – androidguy
    Commented Oct 16, 2014 at 19:59
  • $\begingroup$ @androidguy: We have $g'(a)=\lim_{t\to a}\frac{g(t)-g(a)}{t-a}$. $\endgroup$
    – mathlove
    Commented Oct 16, 2014 at 20:06
  • $\begingroup$ Okay, so it would look like g'(1) = lim t->1 g(1) - g(1) / 1 - 1? $\endgroup$
    – androidguy
    Commented Oct 16, 2014 at 20:17
  • $\begingroup$ @androidguy: No. $g(t)=\frac{|t-2|}{t}$, right? If so, we have$$g'(1)=\lim_{t\to 1}\frac{\frac{|t-2|}{t}-1}{t-1}.$$ $\endgroup$
    – mathlove
    Commented Oct 16, 2014 at 20:19
  • $\begingroup$ Ahh, okay. That makes sense. So then do the same for g'(2)? $\endgroup$
    – androidguy
    Commented Oct 16, 2014 at 20:20

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