Use the definition of the limit to prove the given limit.
Let ε> 0 is any number then we have to find a number δ > 0 so that the following will be true.
| √x - 0| < ε whenever 0 < x - 0 < δ
Or upon a little simplification we need to show,
√x < ε whenever 0 < x < δ
Let's begin with the left hand inequality and illustrates if we can't utilizes that to get a guess for δ . The only simplification which we really have to do here is to square both sides.
√x < ε whenever x < δ2
Hence, it looks like we can chose δ = ε 2 .
Let's verify this. Let ε > 0 be any number and select δ = ε 2 .. Next suppose that 0 < x < ε 2. it gives,
|√x -0|= √x some quick simplification
< (√ ε)2 use the assumption that |x| < ε
= ε simplify
Now we illustrated that,
|√x - 0| = √x whenever 0 < x - 0 < ε 2
and therefore by the definition of the right-hand limit we have,