Question

Calculating limits exactly Use the definition of the derivative to evaluate the following limits. $$\lim _{h \rightarrow 0} \frac{(3+h)^{3+h}-27}{h}$$

Short answer

Answer: The exact limit of the expression as h approaches 0 is 81.

Step-by-step solution

Use the binomial theorem

We will use the binomial theorem to expand the expression \((3+h)^{3+h}\) in order to simplify the limit. The binomial theorem states: $$(x+y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k$$ Plugging in our values, we get: $$(3+h)^{3+h} = \sum_{k=0}^{3+h} \binom{3+h}{k} 3^{3+h-k} h^k$$ Expanding the first few terms, we get: $$(3+h)^{3+h} = \binom{3+h}{0}3^{3+h}h^0 + \binom{3+h}{1}3^{2+h}h^1 + \binom{3+h}{2}3^{1+h}h^2 + \dots$$

Simplify the terms

We will now simplify the terms: $$(3+h)^{3+h} = 3^{3+h} + {(3+h)}3^{2+h}h + \frac{(3+h)(2+h)}{2}3^{1+h}h^2 + \dots$$ Now put this expression back into the limit: $$\lim_{h \rightarrow 0} \frac{3^{3+h} + {(3+h)}3^{2+h}h + \frac{(3+h)(2+h)}{2}3^{1+h}h^2 + \dots -27}{h}$$

Cancel common terms

Now we notice that there's a common term we can cancel, which is \(3^{3+h}\): $$\lim_{h \rightarrow 0} \frac{{(3+h)}3^{2+h}h + \frac{(3+h)(2+h)}{2}3^{1+h}h^2 + \dots -27}{h}$$

Substitute h = 0

Now we can substitute \(h=0\) into the expression, which will give us the final limit value: $$\lim_{h \rightarrow 0} \frac{{(3+h)}3^{2+h}h + \frac{(3+h)(2+h)}{2}3^{1+h}h^2 + \dots -27}{h} = \frac{3\cdot3^{2} + \frac{3\cdot2}{2}3^1\cdot0^2 + \dots -27}{0}$$ However, this will result in division by zero, which is undefined.

Reexamine the expression

Since we have encountered division by zero in Step 4, let us reexamine the expression: $$\lim_{h \rightarrow 0} \frac{3^{3+h} + {(3+h)}3^{2+h}h + \frac{(3+h)(2+h)}{2}3^{1+h}h^2 + \dots -27}{h}$$ Taking a closer look, we notice that when summing up the terms and subtracting 27, the \(3^{3+h}\) term actually cancels out with -27: $$\lim_{h \rightarrow 0} \frac{{(3+h)}3^{2+h}h + \frac{(3+h)(2+h)}{2}3^{1+h}h^2 + \dots}{h}$$

Cancel the h term

Now we can cancel the "h" term from the numerator and denominator: $$\lim_{h \rightarrow 0} ({(3+h)}3^{2+h} + \frac{(3+h)(2+h)}{2}3^{1+h}h + \dots)$$

Substitute h = 0 again

Finally, let's substitute h = 0 again: $$\lim_{h \rightarrow 0} ({(3+h)}3^{2+h} + \frac{(3+h)(2+h)}{2}3^{1+h}h + \dots)= (3\cdot3^{2} + \frac{3\cdot2}{2}3^1\cdot0 + \dots)$$ After evaluating and simplifying, we get: $$3\cdot27=81$$ Therefore, the exact limit is 81: $$\lim _{h \rightarrow 0} \frac{(3+h)^{3+h}-27}{h} = 81$$