How do you evaluate lim_(x->0) (sqrt(4+x)-2)/(3x)?

2 Answers
Mar 14, 2018

Given: lim_(x to 0)(sqrt(4+x)-2)/(3x)

Because the limit yields the indeterminant form 0/0, one should use L'Hôpital's Rule. But, because the topic is "Determining Limits Algebraically", I shall assume that the student has not, yet, learned L'Hôpital's Rule and refrain from using it.

Multiply the expression by 1 in the form of (sqrt(4+x)+2)/(sqrt(4+x)+2):

lim_(x to 0)(sqrt(4+x)-2)/(3x)(sqrt(4+x)+2)/(sqrt(4+x)+2)

The numerator becomes the difference of two squares:

lim_(x to 0)((sqrt(4+x))^2-(2)^2)/((3x)(sqrt(4+x)+2))

Expand the squares:

lim_(x to 0)(4+x-4)/((3x)(sqrt(4+x)+2))

Simplify the numerator:

lim_(x to 0)x/((3x)(sqrt(4+x)+2))

x/x becomes 1:

lim_(x to 0)1/(3(sqrt(4+x)+2))

Now, we may evaluate at x = 0:

1/(3(sqrt(4)+2)) =1/12

This limit is the same as the original expression:

lim_(x to 0)(sqrt(4+x)-2)/(3x) = 1/12

Mar 14, 2018

lim_(x->0) (sqrt(4+x)-2)/(3x) = 1/12

Explanation:

Apply L'Hospital's Rule :

lim_(x->0) (sqrt(4+x)-2)/(3x)

= lim_(x->0) ((d((4+x)^(1/2)-2))/(dx))/((d(3x))/(dx))

=lim_(x->0) (1/2(4+x)^(-1/2))/(3)

=lim_(x->0) 1/(6sqrt(4+x))

=1/(6sqrt(4))

=1/12