Question

Suppose that a and \(\mathrm{b}\) are real positive numbers then the value of \(\lim _{t \rightarrow 0}\left(\frac{b^{t+1}-a^{t+1}}{b-a}\right)^{1 / t}\) has the value equals to (A) $\frac{\mathrm{a} \ln \mathrm{b}-\mathrm{b} \ln \mathrm{a}}{\mathrm{b}-\mathrm{a}}$ (B) $\frac{\mathrm{b} \ln \mathrm{b}-\mathrm{a} \ln \mathrm{a}}{\mathrm{b}-\mathrm{a}}$ (C) \(\mathrm{b} \ln \mathrm{b}-\mathrm{a} \ln \mathrm{a}\) (D) \(\left(\frac{b^{b}}{a^{a}}\right)^{\frac{1}{b-a}}\)

Short answer

Answer: The limit of the expression is (B) \(\frac{b \ln b-a \ln a}{b-a}\).

Step-by-step solution

Rewrite the given expression using logarithms

We will first rewrite the given expression using logarithms. For the expression \(\lim_{t\rightarrow0} (\frac{b^{t+1}-a^{t+1}}{b-a})^{1/t}\), we will take the natural logarithm (ln) of both sides of the expression. This will simplify the expression and make it easier to work with.

Apply Natural Logarithm property on the expression

Now, let's apply ln on both sides of the expression: \(\lim_{t\rightarrow 0} ln(\frac{b^{t+1}-a^{t+1}}{b-a})^{1/t}\). Apply the logarithm property of \(ln(x^p) = p* ln(x)\), we get: $$\lim_{t\rightarrow 0} \frac{1}{t} ln(\frac{b^{t+1}-a^{t+1}}{b-a})$$

Apply Limit property and solve for Limit

We know that \(lim_{x\rightarrow 0} \frac{ln(1+x)}{x} = 1\). Using this property, we will rewrite the given expression as: $$\frac{ln(\frac{b^{t+1}-a^{t+1}}{b-a})}{t} = \frac{ln(1+(\frac{b^{t+1}-a^{t+1}}{b-a}-1))}{t}$$ Now we have it in the form of \(lim_{t \rightarrow 0} \frac{ln(1+x)}{x}\). So, when we take the limit as \(t \rightarrow 0\), we get: $$1*\frac{b\ln b-a\ln a}{b-a} = \frac{b\ln b-a\ln a}{b-a}$$

Compare the result with given choices

Comparing the result with given choices, we can see that our result matches with option (B): $$\frac{b\ln b-a\ln a}{b-a}$$ So, the correct choice is (B) \(\frac{b \ln b-a \ln a}{b-a}\).