Find the limit.

lim
h → 0 square root (49 + h) − 7 / h

Note: Only 49 +h are under square root.

Use L'Hopital's Rule:

sqrt(49+h) = (49+h)^1/2
= 49^1/2 + 1/2 49^-1/2 * h + (1/2)(-1/2)/2 49^-3/2 h^2 ...
= 7 + h/2*7 + ... (higher powers of h)

so,(sqrt(49+h)-7)/h = 1/14 + ... (powers of h, all being 0)

so, limit as h->0 is 1/14

Note that if y=√x, y' = 1/2√x, so y'(7) = 1/14

Ummm. make that Binomial Theorem, not L'Hopital's Rule

To find the limit of the function, we can start by substituting the value of h into the function and see what happens when h gets closer and closer to 0.

The given function is:

f(h) = (sqrt(49 + h) - 7) / h

To evaluate the limit as h approaches 0, we can use algebraic manipulation. We need to get rid of the square root in the numerator since it may cause an indeterminate form when we substitute h = 0.

Let's multiply both the numerator and the denominator of the fraction by the conjugate of the numerator:

f(h) = ((sqrt(49 + h) - 7) / h) * ((sqrt(49 + h) + 7) / (sqrt(49 + h) + 7))

Multiplying these two expressions, we get:

f(h) = (49 + h - 49) / (h * (sqrt(49 + h) + 7))

Now simplifying further:

f(h) = h / (h * (sqrt(49 + h) + 7))

Note that we have an h in both the numerator and the denominator. Now we can cancel out these h terms:

f(h) = 1 / (sqrt(49 + h) + 7)

Now, we can substitute h = 0 into the expression:

lim(h → 0) [1 / (sqrt(49 + h) + 7)]

Now, evaluating the limit:

lim(h → 0) [1 / (sqrt(49 + 0) + 7)]

= 1 / (sqrt(49) + 7)

= 1 / (7 + 7)

= 1 / 14

Therefore, the limit of the given function as h approaches 0 is 1/14.