Question

Evaluate: $\lim_{x \to \infty} \frac{\log 4 + \log (0.5+x)}{\sqrt{x}}$

Answer 1

0

Answer 2

To evaluate the limit $\lim_{x \to \infty} \frac{\log 4 + \log (0.5+x)}{\sqrt{x}}$, we follow these steps:

1. Simplify the numerator: Using the logarithm property $\log A + \log B = \log (AB)$, we can combine the terms in the numerator: $\log 4 + \log (0.5+x) = \log (4 \cdot (0.5+x)) = \log (2+4x)$ So the limit becomes: $\lim_{x \to \infty} \frac{\log (2+4x)}{\sqrt{x}}$

2. Check the indeterminate form: As $x \to \infty$: Numerator: $\log (2+4x) \to \infty$ (since $2+4x \to \infty$) Denominator: $\sqrt{x} \to \infty$ The limit is of the indeterminate form $\frac{\infty}{\infty}$, so we can apply L'Hôpital's Rule.

3. Apply L'Hôpital's Rule (First time): Let $f(x) = \log (2+4x)$ and $g(x) = \sqrt{x}$. Find their derivatives: $f'(x) = \frac{d}{dx} (\log (2+4x)) = \frac{1}{2+4x} \cdot \frac{d}{dx}(2+4x) = \frac{4}{2+4x}$ $g'(x) = \frac{d}{dx} (\sqrt{x}) = \frac{d}{dx} (x^{1/2}) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$ Now, apply L'Hôpital's Rule: $\lim_{x \to \infty} \frac{f'(x)}{g'(x)} = \lim_{x \to \infty} \frac{\frac{4}{2+4x}}{\frac{1}{2\sqrt{x}}}$ $= \lim_{x \to \infty} \frac{4}{2+4x} \cdot 2\sqrt{x}$ $= \lim_{x \to \infty} \frac{8\sqrt{x}}{2+4x}$

4. Evaluate the new limit: The new limit is still of the form $\frac{\infty}{\infty}$. We can either apply L'Hôpital's Rule again or simplify by dividing the numerator and denominator by the highest power of $x$ in the denominator, which is $x$. Dividing numerator and denominator by $x$: $= \lim_{x \to \infty} \frac{\frac{8\sqrt{x}}{x}}{\frac{2}{x}+\frac{4x}{x}}$ $= \lim_{x \to \infty} \frac{\frac{8}{\sqrt{x}}}{\frac{2}{x}+4}$ Now, evaluate the terms as $x \to \infty$: As $x \to \infty$, $\frac{8}{\sqrt{x}} \to 0$. As $x \to \infty$, $\frac{2}{x} \to 0$. So the limit becomes: $\frac{0}{0+4} = \frac{0}{4} = 0$