Question

Question #4a4f9

Answer 1

`1`

Explanation:

All you have to do to solve this is plug in 5 for x.

`lim_(x->5)(|5+2|/(5+2))=(|7|/(7))=1`

If the limiting value was `-2` we would have the following:

as `x-> -2` from the left.

denominator would be negative, numerator would be negative, but because of absolute value we have -(x+2), so numerator is positive.

`lim_(x->-2)(|x+2|/(x+2))= -1`

as `x-> -2` from the right.

denominator would be positive, numerator would be positive.

So:

`lim_(x->-2)(|x+2|/(x+2))= 1`

`lim_(x->-2^-)(|x+2|/(x+2)) != lim_(x->-2^+)(|x+2|/(x+2))`

So:

`lim_(x->-2)(|x+2|/(x+2))` is undefined.

Answer 2

For `x` near `5`, `abs(x+2) = x+2`.

So, for `x` near `5`, we have

`abs(x+2)/(x+2) = (x+2)/(x+2) = 1`.

Therefore,

`lim_(xrarr5)abs(x+2)/(x+2) = lim_(xrarr5)(x+2)/(x+2) = lim_(xrarr5)1 = 1`.