Evaluate as limit approaches 0. (Without using l'hopital's rule.
( Sqrt(4+sin(x))-2 ) / (3x)
lim(√(4+sinx)-2)/(3x)
= lim (cosx/(2√(4+sinx)))/3
-> (1/(2√4))/3 = 1/12
To evaluate the limit as x approaches 0 of (sqrt(4+sin(x))-2)/(3x) without using L'Hopital's rule, we can apply some basic limit rules.
First, let's simplify the expression inside the square root. As x approaches 0, sin(x) approaches 0, so we can approximate sin(x) with just x:
sqrt(4+sin(x)) ≈ sqrt(4+x)
Next, let's simplify the entire expression:
(sqrt(4+x) - 2) / (3x)
Now, we can multiply both the numerator and denominator by the conjugate of the numerator, which is (sqrt(4+x) + 2), to eliminate the square root:
(sqrt(4+x) - 2) / (3x) * (sqrt(4+x) + 2) / (sqrt(4+x) + 2)
This simplifies to:
[(4 + x) - 4] / (3x * (sqrt(4+x) + 2))
x / (3x * (sqrt(4+x) + 2))
1 / (3 * (sqrt(4+x) + 2))
Now, let's take the limit as x approaches 0:
lim(x→0) 1 / (3 * (sqrt(4+x) + 2))
To evaluate this limit, we substitute x = 0 into the expression:
1 / (3 * (sqrt(4+0) + 2))
1 / (3 * (sqrt(4) + 2))
1 / (3 * (2 + 2))
1 / (3 * 4)
1 / 12
Therefore, the limit as x approaches 0 of (sqrt(4+sin(x))-2)/(3x) is equal to 1/12.