Question

(a) Make a conjecture about the value of the limit \(\lim _{k \rightarrow 0} \int_{1}^{b} t^{k-1} d t(b>0)\) (b) Check your conjecture by evaluating the integral and finding the limit. [Hint: Interpret the limit as the definition of the derivative of an exponential function]

Short answer

Answer: The value of the limit of the integral is equal to the natural logarithm of b, or $$\lim _{k \rightarrow 0} \int_{1}^{b} t^{k-1} d t = \ln(b)$$.

Step-by-step solution

Write down the given integral

We are given the following integral: $$\lim _{k \rightarrow 0} \int_{1}^{b} t^{k-1} d t, (b>0)$$

Evaluate the integral

Now let's evaluate the integral. To do this, we will use the power rule for integration, which states that $$\int t^n dt = \frac{t^{n+1}}{n+1} + C$$ Applying the power rule to our integral, we get $$\int_{1}^{b} t^{k-1} dt = \frac{t^k}{k}|_{1}^{b} = \frac{b^k}{k} - \frac{1^k}{k} = \frac{b^k-1}{k}$$

Find the limit

Now we need to find the limit of the evaluated integral as \(k\) approaches 0: $$\lim _{k \rightarrow 0} \frac{b^k-1}{k}$$

Interpret as a derivative and conjecture

Since we are given the hint that we should interpret the limit as the definition of the derivative of an exponential function, let's think about the definition of a derivative: $$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$ If we take the exponential function to be \(f(x) = b^x\), then its derivative \(f'(x)\) would be $$f'(x) = \lim_{h \to 0} \frac{b^{x+h}-b^x}{h}$$ Notice that this resembles our limit $$\lim _{k \rightarrow 0} \frac{b^k-1}{k}$$ If we put \(x=0\) in the expression for \(f'(x)\), we get $$f'(0) = \lim_{h \to 0} \frac{b^h-1}{h}$$ Thus, based on the hint and the similarity between the expressions, our conjecture is that the limit of the integral is equal to the derivative of \(f(x)=b^x\) at \(x=0\), which is \(f'(0)\).

Check the conjecture

We now need to evaluate the derivative of the exponential function \(f(x) = b^x\) at \(x=0\). To do this, we can use the limit definition of the derivative from step 4: $$f'(0) = \lim_{h \to 0} \frac{b^h-1}{h}$$ Now use the L'Hopital's rule (as we have the indeterminate form of the limit) $$f'(0)=\lim_{h \to 0} \frac{\ln(b).b^h.....1}{1}$$ Note here the derivative of \(b^h-1\) w.r.t \(h\) is \(\ln(b).b^h\) $$f'(0) = \ln(b).b^0 = \ln(b)$$ So, the limit of the integral is equal to the derivative of the exponential function at \(x=0\), which is $$\lim _{k \rightarrow 0} \int_{1}^{b} t^{k-1} d t = \ln(b)$$