Question

The value of $\lim_{x \to +\infty} \frac{\ln x}{x^2}$ is given by}

• A. 1
• B. $-\infty$
• C. $+\infty$
• D. 0

Hint:

Logarithmic functions grow much slower than polynomial functions at infinity. Therefore, any limit of the form $\lim_{x\to\infty} \frac{\ln x}{x^a}$ (for $a > 0$) is always 0.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
When evaluating limits at infinity that result in an indeterminate form, we can apply L'Hôpital's Rule.
Key Formula or Approach:
L'Hôpital's Rule states that if $\lim \frac{f(x)}{g(x)}$ yields the indeterminate form $\frac{\infty}{\infty}$, then: \[ \lim \frac{f(x)}{g(x)} = \lim \frac{f'(x)}{g'(x)} \]

Step 2: Detailed Explanation:
Let us evaluate the given limit: \[ L = \lim_{x \to +\infty} \frac{\ln x}{x^2} \]
As $x \to +\infty$, $\ln x \to +\infty$ and $x^2 \to +\infty$, which gives the indeterminate form $\frac{\infty}{\infty}$.
Applying L'Hôpital's Rule by differentiating the numerator and the denominator: \[ f'(x) = \frac{d}{dx}(\ln x) = \frac{1}{x} \] \[ g'(x) = \frac{d}{dx}(x^2) = 2x \]
Now, substitute these derivatives back into the limit: \[ L = \lim_{x \to +\infty} \frac{\frac{1}{x}}{2x} = \lim_{x \to +\infty} \frac{1}{2x^2} \]
As $x \to +\infty$, the denominator $2x^2 \to +\infty$, which means: \[ L = \frac{1}{\infty} = 0 \]

Step 3: Final Answer:
The value of the limit is 0.