Evaluate the limit, if it exists. (3+h)^-1-3^-1/h Solution For the first questio

Question

Evaluate the limit, if it exists.

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

Explanation / Answer

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.

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