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