What is the limit as x approaches 3 of |x+1|+|x-1| ?

The limit as x approaches 3 of |x+1|+|x-1| is 6

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limit as x approaches 3 of |x+1|+|x-1|

x \to 3 lim (∣ x + 1 ∣ + ∣ x - 1 ∣)

6

Solution steps

x \to 3 lim (∣ x + 1 ∣ + ∣ x - 1 ∣)

Plug in the value

x = 3

= ∣ 3 + 1 ∣ + ∣ 3 - 1 ∣

Simplify

∣ 3 + 1 ∣ + ∣ 3 - 1 ∣ : 6

= 6

arcsec (5 x)

\frac{\partial}{\partial L} (L^{\frac{1}{4}} K^{\frac{1}{4}})

\frac{\partial}{\partial x} ((4 x y - 5)^{2})

e^{(\frac{x}{4})}, x = 4, y = 1

\frac{d}{d t} (- 4 sin (t))

What is the limit as x approaches 3 of |x+1|+|x-1| ?
The limit as x approaches 3 of |x+1|+|x-1| is 6