Find the limit. Use l’Hospital’s Rule where appropriate. If there is a more elementary method, consider using it. If l’Hospital’s Rule doesn’t apply, explain wh

Question

Find the limit. Use l’Hospital’s Rule where appropriate. If there is a more elementary method, consider using it. If l’Hospital’s
Rule doesn’t apply, explain why.

 

lim (x --> 0)   [ √(1 + 2x) - √(1 - 4x) ] / x

 

ANSWER: 3

#23, p307

Answer 1

Here's another way to find the limit without using L'Hospital's rule. Leaving off the limit_x->0
until the end.

 

Multiply numerator and denominator by the "conjugate" of the numerator. This results in a difference of squares in the numerator.

 

[√(1+2x) - √(1-4x)] [√(1+2x) + √(1-4x)]

--------------------------------------------------

      x [√(1+2x) + √(1-4x)]

 

This becomes:

 

(1+2x) - (1-4x) = 6x in the numerator.

 

So we have:

 

6x/x { [√(1+2x) + √(1-4x)]}

 

The factor of "x" cancels. so we are taking the limit_x->0 6/[√(1+2x) + √(1-4x)]

 

Substitution of 0 for x results in: 6/2 = 3.

Answer 2

This is similar to your other problem involving L'Hopital's rule.

 

Taking the limit initially, you would get 0/0 therefore we could apply L'Hopital's:

 

The derivative of the numerator: 1/2(1+2x)^-1/2(2) -1/2(1-4x)^-1/2(-4)

The derivative of the denominator (x) simply becomes 1.

 

Since the numerator is continuous at 0, simply plug in 0 and get 1 +2 = 3

(the roots simply equal 1 and become irrelevant, the only numbers that come into play are the 1/2 you get from applying the power rule when taking the derivative and the (2) and (-4) at the ends of each term you get by applying the chain rule).