Evaluate the following limit, if it exists:
lim h-->0 ((sqrt9+h)-3/(h)
if f(x) = √x
the limit is f'(9)
f'(x) = 1/(2√x)
f'(9) = 1/6
If you must evaluate the limit, just use the binomial theorem:
(9+h)^(1/2) = 9^(1/2) + (1/2)/1! 9^(-1/2)h + (1/2)(-1/2)/2! 9^(-3/2) h^2 + ...
subtract 3, divide by h, and set h=0
To evaluate the limit, we can simplify the expression first:
Since we have a square root term in the numerator, we can multiply the expression by the conjugate of the numerator to eliminate the square root. The conjugate of a term a + b is a - b.
So, multiplying the expression by the conjugate of the numerator gives us:
((sqrt(9+h) - 3) * (sqrt(9+h) + 3)) / (h * (sqrt(9+h) + 3))
Expanding the numerator using the difference of squares, we have:
((9 + h) - 3^2) / (h * (sqrt(9+h) + 3))
Simplifying further, we get:
(h + 6) / (h * (sqrt(9+h) + 3))
Now, we can simplify the fraction by factoring out h from the numerator:
h * (1 + 6/h) / (h * (sqrt(9+h) + 3))
Next, we can cancel out the h terms in the numerator and denominator:
(1 + 6/h) / (sqrt(9+h) + 3)
Now, we can evaluate the limit by substituting h with 0:
As h approaches 0, the fraction becomes:
(1 + 6/0) / (sqrt(9+0) + 3)
However, since there is a division by zero in the expression, the limit does not exist.