limits existing

Question

Evualate the limit if it exists

lim x→0 (√(1+x) - √(1-x))/ (∣x ∣) 

Answer 1

What Alex says about the right and left limit needing to exist and be equal is correct.

There are two ways to see the limiting behavior of the numerator:

1)note that sqrt(1+x) approaches 1 + 1/2 x, so sqrt(1+x) - sqrt(1-x) approaches (1+1/2x) - (1 - 1/2 x) = x; 2)multiply the numerator by its conjugate:

(sqrt(1+x)-sqrt(1-x)) * (sqrt(1+x)+sqrt(1-x))/(sqrt(1+x)+sqrt(1-x)) = (1+x-(1-x))/(sqrt(1+x)+sqrt(1-x)) = 2x/(sqrt(1+x)+sqrt(1-x)), which approaches 2x/2 = x

Now, if we consider the denominator so that we can get the right and left limit as x approaches 0, x/|x| = 1 for positive x and -1 for negative x. Thus, there is no limit.

If you plug in small values for x, as you should do on limit problems, you can easily verify this result.

Answer 2

Call f(x) = (√(1+x) - √(1-x))/|x|.

Claim:

The limit of f(x) as x approaches 0 exists

if and only if

both

lim f(x) as x approaches 0 from the left exists

And

lim f(x) as x approaches 0 from the right exists

And

these two limits are equal.

So consider the limit of the function as it approaches both sides of zero.

This way you can use the piece-wise definition of |x| to remove the absolute values.

Recall a piece-wise definition |x| = { x if x>0 , -x if x<0, and 0 if x = 0.

Then you can just use L’Hopital’s rule (B/c I believe direct evaluation leads to the indeterminate case of 0/0) to find the lim of f(x) as x approaches 0 from one side of zero. Then do the process again to find the lim f(x) as x approaches 0 from the other side of zero. If they both exist and equal each other, then that’s the limit. If not the limit doesn’t exist.

********************

Tom’s way works too. Isn’t it cool to see more than one path! I’ve always thought the art in math is in the path! Anyway let’s summarize the approaches.

DEFINITIONS USED

f(x) = (√(1+x) - √(1-x))/|x|

|x| = { x if x≥0,

-x if x<0.

THEOREMS UTILIZED

1) The limit of a function k(x) as x approaches some value a exists and is L, i.e.

lim_x→a k(x) = L,

if and only if

lim_x→a¯ k(x) = L,

and

lim_x→a^+ k(x) = L.

2) Given k(x) = s(x) / t(x) is differentiable on the interval of interest where

lim_x→a s(x) = 0,

and

lim_x→a t(x) = 0.

Then L’Hopital’s Rule states that lim_x→a k(x) = lim_x→a s’(x) / t’(x).

METHOD USING CLEVER MULTIPLICATION BY 1 (CONJUGATE)

lim_x→0 f(x) = lim_x→0 (√(1+x) - √(1-x))/|x| * (√(1+x) + √(1-x))/(√(1+x) + √(1-x))

= lim_x→0 ((1+x) - (1-x))/(|x|(√(1+x) + √(1-x)))

= lim_x→0 2x/(|x|(√(1+x) + √(1-x))).

Now consider limits approaching 0 from the left and right.

Left

lim_x→0¯ f(x) = lim_x→0¯ 2x/((-x)(√(1+x) + √(1-x))) Using definition of |x| for x<0.

= - lim_x→0¯ 2/(√(1+x) + √(1-x))

= - 2/(√(1) + √(1))

= - 2/2

= - 1.

Right

lim_x→0^+ f(x) = lim_x→0^+ 2x/((x)(√(1+x) + √(1-x))) Using definition of |x| for x>0.

= lim_x→0¯ 2/(√(1+x) + √(1-x))

= 2/(√(1) + √(1))

= 2/2

= 1.

Hence,

lim_x→0¯ f(x) = -1

and

lim_x→0^+ f(x) = 1.

Thus the limit of f(x) as x approaches 0 does not exist.

METHOD USING L’HÔPITAL’S RULE

For this method, let’s start by considering the limits as f(x) approaches 0 from the left and the right.

Left

lim_x→0¯ f(x) = lim_x→0¯ (√(1+x) - √(1-x))/(-x) Using the definition of |x| above for x<0

= - lim_x→0¯ (√(1+x) - √(1-x))/x.

Right

lim_x→0^+ f(x) = lim_x→0^+ (√(1+x) - √(1-x))/(x) Using the definition of |x| above for x>0

= lim_x→0^+ (√(1+x) - √(1-x))/x.

NOW DEFINE

j(x) = (√(1+x) - √(1-x))/x.

Then

Left

lim_x→0¯ f(x) = - lim_x→0¯ j(x)

and

Right

lim_x→0^+ f(x) = lim_x→0^+ j(x).

Next consider

lim_x→0 j(x) = lim_x→0 (√(1+x) - √(1-x))/x.

Moreover if we DEFINE

g(x) = (√(1+x) - √(1-x))

and

h(x) = x.

Notice that lim_x→0 g(x) = 0 = lim_x→0 h(x),

and

g’(x) = 1/(2√(1+x)) - 1/(2√(1-x))(-1)

= 1/(2√(1+x)) + 1/(2√(1-x)),

h‘(x) = 1.

So by L’Hopital’s Rule,

lim_x→0 j(x) = lim_x→0 g’(x) / h’(x)

= lim_x→0 1/(2√(1+x)) + 1/(2√(1-x))

= 1/(2√(1)) + 1/(2√(1))

= 1.

Therefore,

Left

lim_x→0¯ f(x) = - lim_x→0¯ j(x) = -1

and

Right

lim_x→0^+ f(x) = lim_x→0^+ j(x) = 1.

Hence it must be that lim_x→0 f(x) does not exist.