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

Question

Aurora Alvarado · Accepted Answer

Does not exist.

#lim(sqrt(3+x)-(sqrt(3))/x),x,0# You can think of this limit in two parts: first, as a limit goes to zero of #sqrt(3+x)# and second as a limit goes to zero of #(sqrt(3))/x#. It is clear that limit for the first part would be just #sqrt(3)#. Since, the #sqrt(3)# won't have any significant contribution to the limit, we can easily modify the question from #lim(sqrt(3+x)-(sqrt(3))/x)# to #lim(-sqrt(3)/x)#. Now, we can find the limit of the function #-sqrt(3)/x# as x approaches zero from negative side and we note that the smaller and smaller negative number that is very close to zero but it's not zero would produce really big numbers. So, the limit as x goes to zero from negative direction is positive infinity. Similarly, the limit as x goes to zero from positive direction is negative infinity. Since the limits from left and right directions are not the same, the limit of this function is said to be does not exist.

Isabella Ackerman · Answer

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

First, let's simplify the expression by rationalizing the denominator. Multiply both the numerator and denominator by the conjugate of the denominator, which is sqrt(3+x) + sqrt(3).

This gives us [(sqrt(3+x) - sqrt(3))(sqrt(3+x) + sqrt(3))]/[(x)(sqrt(3+x) + sqrt(3))].

Expanding the numerator using the difference of squares, we get [(3+x) - 3]/[(x)(sqrt(3+x) + sqrt(3))].

Simplifying further, we have x/[(x)(sqrt(3+x) + sqrt(3))].

Now, we can cancel out the x terms in the numerator and denominator, giving us 1/(sqrt(3+x) + sqrt(3)).

As x approaches 0, the expression sqrt(3+x) approaches sqrt(3), so the denominator becomes 2(sqrt(3)).

Therefore, the limit of sqrt(3+x) - sqrt(3)/x as x approaches 0 is 1/(2(sqrt(3))).