Question

$\lim _{y \rightarrow 0}\left[\lim _{x \rightarrow \infty} \frac{\exp \left(\mathrm{x} \ell \mathrm{n}\left(1+\frac{\mathrm{ay}}{\mathrm{x}}\right)\right)-\exp \left(\mathrm{x} \& \mathrm{n}\left(1+\frac{\mathrm{by}}{\mathrm{x}}\right)\right)}{\mathrm{y}}\right]$ $\begin{array}{ll}\text { (A) } \mathrm{a}+\mathrm{b} & \text { (B) } \mathrm{a}-\mathrm{b}\end{array}$ (C) \(b-a \quad\) (D) \(-(a+b)\)

Short answer

Question: Find the limit $$\lim_{y \rightarrow 0}\left[\lim_{x \rightarrow \infty} \frac{\exp \left(\mathrm{x}\;\mathrm{ln}\left(1+\frac{\mathrm{ay}}{\mathrm{x}}\right)\right)-\exp \left(\mathrm{x}\;\mathrm{ln}\left(1+\frac{\mathrm{by}}{\mathrm{x}}\right)\right)}{\mathrm{y}}\right]$$ Answer: (B) \(a - b\)

Step-by-step solution

Simplify the expression using logarithm properties

First, let's simplify the expression using logarithm properties: $$\lim_{y \rightarrow 0}\left[\lim_{x \rightarrow \infty} \frac{\exp \left(\mathrm{x}\;\mathrm{ln}\left(1+\frac{\mathrm{ay}}{\mathrm{x}}\right)\right)-\exp \left(\mathrm{x}\;\mathrm{ln}\left(1+\frac{\mathrm{by}}{\mathrm{x}}\right)\right)}{\mathrm{y}}\right]$$ Since the base of the natural logarithm is a common base for both exponentials, we have: $$\lim_{y \rightarrow 0}\left[\lim_{x \rightarrow \infty} \frac{\left(1+\frac{\mathrm{ay}}{\mathrm{x}}\right)^{\mathrm{x}}-\left(1+\frac{\mathrm{by}}{\mathrm{x}}\right)^{\mathrm{x}}}{\mathrm{y}}\right]$$

Apply L'Hopital's rule to the limit of x

Now let's apply L'Hospital's rule to the inner limit: $$\lim_{x \rightarrow \infty} \frac{\left(1+\frac{\mathrm{ay}}{\mathrm{x}}\right)^{\mathrm{x}}-\left(1+\frac{\mathrm{by}}{\mathrm{x}}\right)^{\mathrm{x}}}{\mathrm{y}}$$ Differentiate the numerator and denominator with respect to x, and simplify: $$\lim_{x \rightarrow \infty} \frac{a\left(1+\frac{\mathrm{ay}}{\mathrm{x}}\right)^{\mathrm{x}-1}-b\left(1+\frac{\mathrm{by}}{\mathrm{x}}\right)^{\mathrm{x}-1}}{0}$$ Now we need to let x approach infinity and see what happens to the expression: $$\lim_{x \rightarrow \infty}\left(\frac{a\left(1+\frac{\mathrm{ay}}{\mathrm{x}}\right)^{\mathrm{x}-1}-b\left(1+\frac{\mathrm{by}}{\mathrm{x}}\right)^{\mathrm{x}-1}}{0}\right)=\frac{\infty-\infty}{0}$$ This is an indeterminate form, so we need to apply L'Hopital's rule again to the limit of x: $$\lim_{x \rightarrow \infty} \frac{a\left(1+\frac{\mathrm{ay}}{\mathrm{x}}\right)^{\mathrm{x}-2}-b\left(1+\frac{\mathrm{by}}{\mathrm{x}}\right)^{\mathrm{x}-2}}{0}$$ Now let x approach infinity: $$\lim_{x \rightarrow \infty}\left(\frac{a\left(1+\frac{\mathrm{ay}}{\mathrm{x}}\right)^{\mathrm{x}-2}-b\left(1+\frac{\mathrm{by}}{\mathrm{x}}\right)^{\mathrm{x}-2}}{0}\right)=\frac{\infty-\infty}{0}$$ This is still an indeterminate form. Repeat L'Hopital's rule until the indeterminate form disappears: $$\lim_{x \rightarrow \infty} \frac{a\left(1+\frac{\mathrm{ay}}{\mathrm{x}}\right)^{0}-b\left(1+\frac{\mathrm{by}}{\mathrm{x}}\right)^{0}}{0}$$ $$\lim_{x \rightarrow \infty}\left(\frac{a - b}{0}\right) = a - b$$ So we find that the inner limit equals \(a - b\).

Find the final limit

Now we can substitute the inner limit value back into the original expression: $$\lim_{y\rightarrow 0}(a-b)$$ As there is only a constant value in this limit, we can simply report the answer as: \(a - b\). The correct answer is (B).