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

Apply L'Hospital's Rule :

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The numerator becomes the difference of two squares:

Expand the squares:

Simplify the numerator:

This limit is the same as the original expression:

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To evaluate the limit lim_(x->0) (sqrt(4+x)-2)/(3x), we can use algebraic manipulation and the limit properties. First, we can simplify the expression by multiplying the numerator and denominator by the conjugate of the numerator, which is sqrt(4+x) + 2. This will help us eliminate the square root in the numerator.

After multiplying, we get (sqrt(4+x) - 2)(sqrt(4+x) + 2)/(3x)(sqrt(4+x) + 2). Simplifying further, we have (4+x - 4)/(3x)(sqrt(4+x) + 2). The numerator cancels out, leaving us with x/(3x)(sqrt(4+x) + 2).

Next, we can simplify the expression by canceling out the common factor of x in the numerator and denominator. This gives us 1/(3(sqrt(4+x) + 2)).

Now, we can evaluate the limit as x approaches 0. Plugging in 0 for x, we get 1/(3(sqrt(4+0) + 2)) = 1/(3(2)) = 1/6.

Therefore, the limit of (sqrt(4+x)-2)/(3x) as x approaches 0 is 1/6.

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