# Thread: Absolute value of the limit of a function is equal to the absolute value of the limit

1. ## Absolute value of the limit of a function is equal to the absolute value of the limit

The question:

Let f: D-> R and let c be an accumulation point of D. Suppose that lim_x->c f(x)=L.

Prove that lim_x->c |f(x)|=|L|.

A sketch of what I have:
Definition of limit and then

|f(x)-L|<$\displaystyle \varepsilon$

||f(x)|-|L|| $\displaystyle \leq$ |f(x)-L|<$\displaystyle \varepsilon$

But I'm sketchy on the reasons and if this is the right direction. Any ideas? Thanks.

2. ## Re: Absolute value of the limit of a function is equal to the absolute value of the l

Originally Posted by CountingPenguins
The question:

Let f: D-> R and let c be an accumulation point of D. Suppose that lim_x->c f(x)=L.

Prove that lim_x->c |f(x)|=|L|.

A sketch of what I have:
Definition of limit and then

|f(x)-L|<$\displaystyle \varepsilon$

||f(x)|-|L|| $\displaystyle \leq$ |f(x)-L|<$\displaystyle \varepsilon$

But I'm sketchy on the reasons and if this is the right direction. Any ideas? Thanks.
Just throw in those $\displaystyle \delta$s from the definition of limit and you have your proof