Using the definition of a derivative, solve $\lim_{x \to 0} \frac{(2+h)^{3+h} - 8}{h}$

Question

I can't figure out where to even start, I have looked up the answer on Desmos but it uses L'Hopitals rule which i haven't learned yet.

$$\lim_{h \to 0} \frac{(2+h)^{3+h} - 8}{h}$$ I see that i can use the log rules to rewrite it as $$\lim_{h \to 0} \frac{e^{(3+h)\ln(2+h)} - 8}{h}$$ but that only confuses me more.

Answer 1

Per Levent's comment, supposed that you're given $f(x) = x^{x+1}$. By definition of the derivative:

$$f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \rightarrow 0} \frac{(x+h)^{x+1+h} - x^{x+1}}{h}$$

Now, plug in $x=2$.

$$f'(2) = \lim_{h \rightarrow 0} \frac{(2+h)^{3+h} - 8}{h}$$

Hey, that's conveniently familiar! Now your question can be rephrased as “Find $f'(2)$.”

Now, what's the derivative of $y = f(x) = x^{x+1}$? The equation may be easier to work with if you take logarithms.

$$ \ln y = (x + 1) \ln x$$

You can then use implicit differentiation to get:

$$\frac{1}{y} \frac{dy}{dx} = \ln x + \frac{x+1}{x} = \ln x + 1 + \frac{1}{x}$$

$$\frac{dy}{dx} = y(\ln x + 1 + \frac{1}{x})$$

And plugging in $x=2$ and $y=8$ gives:

$$f'(2) = 8(\ln 2 + 1 + \frac{1}{2}) = 8 \ln 2 + 12$$

Answer 2

I follow up on your suggestion to use a logarithm. Define

$$ \phi(h)= e^{(3+h)\ln(2+h)}, \text{ noting that } \phi(0)= 8.$$

On the one hand, $$\phi'(0) \;=\; \lim_{h \to 0}\frac{\phi(h)-\phi(0)}{h-0},$$ which is exactly your target.

But applying the basic rules for derivatives of complicated expressions:

$$ \phi'(h)= \phi(h)\left[\ln(2+h) + (3+h)\frac1{2+h}\right],$$

$$ \phi'(0)= 12+8\ln 2.$$