# Find `lim_(x->2)|x-5|/(x-2)`

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The limit `lim_(x->2)|x-5|/(x-2)` has to be determined.

`lim_(x->2)|x-5|/(x-2)`

Substituting x = 2 gives `|-3|/0`

But any number divided by 0 is infinity.

**The limit **`lim_(x->2)|x-5|/(x-2) = oo`

The limit `lim_(x->2) (|x - 5|)/(x - 2)` is required.

|x - 5| is the absolute value of (x - 5).

In this problem it is possible for x to approach 2 from two directions, values of x greater than 2 and values of x less than 2. When `x<= 2` , |(x - 5)| `>= 0` and `x - 2 <= 0` .

The limit `lim_(x->2-)(|x-5|)/(x-2) = -oo`

When x approaches 2 from the right hand side where values are `>=2` , `|(x - 5)| >= 0` and `(x - 2) >= 0`

The limit `lim_(x->2+)(|x-5|)/(x-2) = oo`

The following graph visually confirms the result.