Question

Consider the function \(f(x)=\left(\frac{a x+1}{b x+2}\right)^{x}\) where \(a^{2}+b^{2} \neq 0\) then \(\lim f(x)\) (A) exists for all values of \(a\) and \(b\) (B) is zero for \(\mathrm{a}<\mathrm{b}\) (C) is non existent for \(\mathrm{a}>\mathrm{b}\) (D) is e \({ }^{-\left(\frac{1}{a}\right)}\) or \(e^{-\left(\frac{1}{b}\right)}\) if \(a=b\)

Short answer

Question: Analyze the behavior of the limit of the function \(f(x) = \left(\frac{a x+1}{b x+2}\right)^{x}\) as \(x\) approaches infinity depending on the values of \(a\) and \(b\). Answer: The limit of the function \(f(x)\) as \(x\) approaches infinity is 0 when \(ab\).

Step-by-step solution

Interpret the Limit

We need to find the limit of the function as x approaches infinity, which can be represented as follows: $$\lim_{x\to\infty}f(x) = \lim_{x\to\infty}\left(\frac{a x+1}{b x+2}\right)^{x}$$

Divide the terms of the fraction by x

To work with this limit, we will first divide the terms of the fraction by x. The resulting expression is then: $$\lim_{x\to\infty}\left(\frac{\frac{a x}{x}+\frac{1}{x}}{\frac{b x}{x}+\frac{2}{x}}\right)^{x}$$

Simplify the expression

Next, we will simplify the expression inside the limit as follows: $$\lim_{x\to\infty}\left(\frac{a+\frac{1}{x}}{b+\frac{2}{x}}\right)^{x}$$

Analyzing the limit

As \(x\) approaches infinity, the function behaves like the following limit: $$\lim_{x\to\infty}\left(\frac{a}{b}\right)^{x} = \lim_{x\to\infty}e^{x\ln\left(\frac{a}{b}\right)} $$ Now, let's examine the cases mentioned in the given options.

a < b case (Option B)

If \(a<b\), then the fraction \(\frac{a}{b}\) is less than 1. Let \(r=\frac{a}{b}\), then \(0<r<1\). As \(x\) approaches infinity, we have: $$\lim_{x\to\infty}e^{x\ln r} = e^{\ln r^\infty} = e^{-\infty}=0$$ Since the limit converges to 0 in this case, option (B) is correct.

a > b case (Option C)

If \(a>b\), then the fraction \(\frac{a}{b}\) is greater than 1. In this case, as \(x\) approaches infinity, the limit of the function does not exist since the function increases without bound (goes to infinity). Therefore, in this case option (C) is correct.

a = b case (Option D)

If \(a=b\), then the fraction \(\frac{a}{b}\) is equal to 1, and the limit becomes: $$\lim_{x\to\infty} \left(\frac{a}{b}\right)^{x} = \lim_{x\to\infty} e^{x\ln 1} = e^0 = 1$$ However, this result does not match either value mentioned in option (D). Thus, option (D) is incorrect.

Conclusion

Based on the analysis, we can conclude that the options (B) and (C) are correct regarding the limit of the function \(f(x)\) depending on the values of \(a\) and \(b\): \(\lim_{x\to\infty} f(x) = 0\ \mathrm{for}\ ab\).