Find the limit (if it exists) as h --> 0 of ((3 + h)^-1 - 3^-1) / h ?

For the first question, re-write the negative exponents as a fraction to give:

lim (h-->0) [(3 + h)^(-1) - 3^(-1)]/h

= lim (h-->0) [1/(3 + h) - 1/3]/h.

Multiplying the numerator and denominator by 3(3 + h) (the LCD of the fractions in the numerator) yields:

lim (h-->0) [1/(3 + h) - 1/3]/h

= lim (h-->0) [3 - (3 + h)]/[3h(3 + h)]

= lim (h-->0) -h/[3h(3 + h)]

= lim (h-->0) -1/[3(3 + h)], by canceling h

= -1/[3 * (3 + 0)]

= -1/9.

The second problem is unclear. Is this 1/t - 1/t^2 + 2, (1/t - 1)/(t^2 + 2), or 1/t - 1/(t^2 + 2)? If we take expression literally, 1/t - 1/t^2 + 2 --> -infinity as t --> 0. In other words, the limit doesn't exist.

I hope this helps!

By the definition of the derivative, (1) is the derivative of the function 1/x at x = 3.

d/dx(1/x) = -1/x^2

At x = 3, this becomes -1/9.

(2)

= lim (t - 1)/t^2 + 2

= lim (1 - 1/t)/t^3 + 2

As t approaches 0, 1/t approaches 0 and 1/t^3 diverges. So, the numerator converges and therefore the function diverges.

limit((1/(3+h)-3^(-1))/h, h = 0) = - 1/9

limit(1/t-1/t^2+2, t = 0) = - infinity

((1/(3 + h)) - (1/3)) / h

= ((3/(3(3 + h))) - ((3 + h)/(3(3 + h)))) / h

= ((3 - (3 + h)) / (3(3 + h))) / h

= ((3 - 3 - h) / (9 + h)) / h

= ((-h) / (9 + h)) / h

= -1 / (9 + h)

If h approaches zero, we have:

= -1 / 9