Question

Calculating limits exactly Use the definition of the derivative to evaluate the following limits. $$\lim _{h \rightarrow 0} \frac{\ln \left(e^{8}+h\right)-8}{h}$$

Short answer

Answer: The value of the limit is $$\frac{1}{e^8}$$.

Step-by-step solution

Identify the function

In this case, the function is given by the natural logarithm, i.e., $$f(x) = \ln(x)$$. The limit is set up like the definition of the derivative, with the function being evaluated at $$e^8$$, an expression of the form $$f(a + h) - f(a)$$.

Simplify the given limit using logarithmic properties

We can simplify the given limit using the logarithmic property of the natural logarithm, which states that $$\ln(e^k) = k$$ for any $$k$$: $$\lim _{h \rightarrow 0} \frac{\ln \left(e^{8}+h\right)-8}{h}$$

Apply the derivative definition to the natural logarithm

The definition of the derivative with respect to $$x$$ of the function $$f(x) = \ln(x)$$ is as follows: $$f'(x) = \lim_{h \rightarrow 0} \frac{\ln(x + h) - \ln(x)}{h}$$ In this case, $$x = e^8$$, so the limit we need to compute is: $$\lim _{h \rightarrow 0} \frac{\ln \left(e^{8}+h\right)-\ln(e^8)}{h}$$

Use the logarithmic property of difference

We can further simplify using the logarithmic property that states $$\ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right)$$: $$\lim _{h \rightarrow 0} \frac{\ln \left(\frac{e^{8}+h}{e^8}\right)}{h}$$

Simplify and apply limit properties

Now, we can factor out a $$\frac{1}{e^8}$$ from the argument and simplify: $$\lim _{h \rightarrow 0} \frac{\ln \left(1+\frac{h}{e^8}\right)}{h}$$ We now have a limit of the form $$\lim_{x \rightarrow 0} \frac{\ln(1+x)}{x}$$. Let $$x = \frac{h}{e^8}$$, then we have: $$\lim _{x \rightarrow 0} \frac{\ln \left(1+x\right)}{x}$$

Evaluate the limit

The limit we have now is a well-known limit in calculus, which is equal to $$1$$: $$\lim _{x \rightarrow 0} \frac{\ln(1+x)}{x} = 1$$

Compute the final answer

Since we had made a substitution of $$x = \frac{h}{e^8}$$, we need to account for this to find the original limit. To do this, we simply multiply the result by the coefficient of $$h$$ in our substitution, which is $$\frac{1}{e^8}$$: $$\lim _{h \rightarrow 0} \frac{\ln \left(e^{8}+h\right)-8}{h} = \frac{1}{e^8}$$ Therefore, the final answer is: $$\lim _{h \rightarrow 0} \frac{\ln \left(e^{8}+h\right)-8}{h} = \frac{1}{e^8}$$