How do you find the limit of #(sqrt(x+3) - sqrt(3)) / x# as x approaches 0?

Question

Ezekiel Alexander · Accepted Answer

#" The Reqd. Lim.="sqrt3/6#.

The Reqd. Limit#=lim_(xrarr0)(sqrt(x+3)-sqrt3)/x# #=lim_(xrarr0) (sqrt(x+3)-sqrt3)/x xx (sqrt(x+3)+sqrt3)/(sqrt(x+3)+sqrt3)# #=lim_(xrarr0) {(sqrt(x+3)^2-sqrt3^2)}/{x(sqrt(x+3)+sqrt3)}# #=lim_(xrarr0) (x+3-3)/{x(sqrt(x+3)+sqrt3)}# #=lim_(xrarr0) 1/((sqrt(x+3)+sqrt3)# #=1/(sqrt3+sqrt3)=1/(2sqrt3)=sqrt3/6.# #:." The Reqd. Lim.="sqrt3/6#.

Charles Anctil · Answer

undefined To find the limit of (sqrt(x+3) - sqrt(3)) / x as x approaches 0, we can use algebraic manipulation and the concept of limits.

First, we rationalize the numerator by multiplying both the numerator and denominator by the conjugate of the numerator, which is (sqrt(x+3) + sqrt(3)). This gives us:

[(sqrt(x+3) - sqrt(3)) / x] * [(sqrt(x+3) + sqrt(3)) / (sqrt(x+3) + sqrt(3))]

Simplifying this expression, we get:

[(x+3) - 3] / (x * (sqrt(x+3) + sqrt(3)))

Further simplifying, we have:

x / (x * (sqrt(x+3) + sqrt(3)))

Canceling out the x terms, we are left with:

1 / (sqrt(x+3) + sqrt(3))

Now, as x approaches 0, the expression inside the square root, (x+3), approaches 3. Therefore, the limit of the denominator as x approaches 0 is 2 * sqrt(3).

Substituting this value back into our expression, we get:

1 / (2 * sqrt(3) + sqrt(3))

Simplifying, we have:

1 / (3 * sqrt(3))

Thus, the limit of (sqrt(x+3) - sqrt(3)) / x as x approaches 0 is 1 / (3 * sqrt(3)).