91Ó°ÊÓ

Before applying L'Hôpital's Rule, we need to check if the given limit goes to an indeterminate form. Let's check whether the numerator and the denominator both tend to infinity when x tends to infinity. For the numerator: As x goes to infinity, both 3x and \(5e^x\) also go to infinity. Hence, the whole numerator ln(3x+\(5e^x\)) also goes to infinity. For the denominator: As x goes to infinity, both 7x and \(3e^{2x}\) go to infinity. Hence, the whole denominator, ln(7x+\(3e^{2x}\)), goes to infinity as well. Since both the numerator and the denominator go to infinity as x goes to infinity, the limit has an indeterminate form of the type ∞/∞.

Since we have an indeterminate form of the type ∞/∞, we can apply L'Hôpital's Rule. To do this, we differentiate the numerator and the denominator of the function separately. The derivatives of the numerator and the denominator are as follows: Numerator: $$\frac{d}{dx} \ln(3x + 5e^x) = \frac{1}{3x+5e^x} \cdot (3 + 5e^x) = \frac{3+5e^x}{3x+5e^x}$$ Denominator: $$\frac{d}{dx} \ln(7x + 3e^{2x}) = \frac{1}{7x+3e^{2x}} \cdot (7 + 6e^{2x}) = \frac{7+6e^{2x}}{7x+3e^{2x}}$$ Now, the limit we need to evaluate becomes: $$\lim _{x \rightarrow \infty} \frac{\frac{3+5e^x}{3x+5e^x}}{\frac{7+6e^{2x}}{7x+3e^{2x}}}$$

Let's simplify the expression above by inverting the denominator and multiplying it with the numerator: $$\lim _{x \rightarrow \infty} \frac{(3 + 5 e^x)(7x + 3 e^{2x})}{(3x + 5 e^x)(7 + 6 e^{2x})}$$ Now, let's find the highest power of x in the numerator and the denominator in the given expression. In this case, it is \(e^x\) for the numerator and \(e^{2x}\) for the denominator. Let's divide both the numerator and the denominator by \(e^x\): $$\lim _{x \rightarrow \infty} \frac{(3e^{-x} + 5)(7x + 3 e^{x})}{(3x + 5)(7e^{-x} + 6 e^{x})}$$ Clear out the terms that tend to zero: $$\lim _{x \rightarrow \infty} \frac{5(7x + 3 e^{x})}{(3x + 5)(6 e^{x})}$$ Again, we have another indeterminate form. Hence, we will apply L'Hôpital's Rule one more time. The derivatives of the numerator and the denominator are as follows: Numerator: $$\frac{d}{dx} (5(7x + 3e^x)) = 5(7 + 3e^x)$$ Denominator: $$\frac{d}{dx} ((3x + 5)(6e^x)) = (3(6e^x) + 6e^x(3x + 5))$$ Now, the limit we need to evaluate becomes: $$\lim_{x \rightarrow \infty} \frac{5(7 + 3e^x)}{3(6e^x) + 6e^x(3x + 5)}$$ We again have an indeterminate form here. Let's simplify the expression by dividing both the numerator and the denominator by \(e^x\): $$\lim _{x \rightarrow \infty} \frac{5(7e^{-x} + 3)}{3(6) + 6(3x)}$$ Now, clear out the terms that tend to zero: $$\lim_{x \rightarrow \infty} \frac{5(3)}{6(3x)}$$ Finally, we can find the limit: $$\lim_{x \rightarrow \infty} \frac{5(3)}{6(3x)} = \frac{5}{6x}$$ As x goes to infinity, the limit of the given expression is 0.