Question

Evaluate \[ \lim_{x\to 0}\frac{\log(4+x)^x-\log 4^x}{\sin^2 x}. \]

• A. \(4\)
• B. \(\dfrac14\)
• C. \(2\)
• D. \(\dfrac12\)

Hint:

For limits involving logarithms, first convert powers using \(\log(a^x)=x\log a\). Then try to create the standard limit \[ \lim_{t\to0}\frac{\log(1+t)}{t}=1. \] Whenever \(\sin x\) appears in the denominator, use the approximation \(\sin x \sim x\) near \(x=0\).

Answer 1

The Correct Option is B

Concept: This problem involves the use of logarithmic identities together with standard limits from differential calculus. The two key results used are: \[ \log(a^x)=x\log a, \] and \[ \lim_{t\to 0}\frac{\log(1+t)}{t}=1. \] Also, \[ \lim_{x\to 0}\frac{\sin x}{x}=1, \] which implies \[ \sin^2 x \sim x^2 \quad \text{as} \quad x\to 0. \] Using these facts, the given limit can be transformed into a standard form.

Step 1: Simplify the logarithmic expression. Using the identity \[ \log(a^x)=x\log a, \] we obtain \[ \log(4+x)^x = x\log(4+x), \] and \[ \log 4^x = x\log 4. \] Hence the numerator becomes \[ x\log(4+x)-x\log4. \] Taking \(x\) common, \[ x\Bigl(\log(4+x)-\log4\Bigr). \] Using \[ \log A-\log B = \log\left(\frac{A}{B}\right), \] we get \[ x\log\left(\frac{4+x}{4}\right). \] Thus, \[ x\log\left(1+\frac{x}{4}\right). \] Therefore the limit becomes \[ \lim_{x\to0} \frac{x\log\left(1+\frac{x}{4}\right)} {\sin^2x}. \]

Step 2: Replace \(\sin^2x\) using the standard approximation. Since \[ \sin x \sim x \quad \text{as} \quad x\to0, \] we have \[ \sin^2x \sim x^2. \] Therefore, \[ \lim_{x\to0} \frac{x\log\left(1+\frac{x}{4}\right)} {\sin^2x} = \lim_{x\to0} \frac{x\log\left(1+\frac{x}{4}\right)} {x^2}. \] Cancelling one factor of \(x\), \[ = \lim_{x\to0} \frac{\log\left(1+\frac{x}{4}\right)} {x}. \]

Step 3: Convert into a standard logarithmic limit. Let \[ t=\frac{x}{4}. \] Then \[ x=4t. \] Substituting, \[ \frac{\log\left(1+\frac{x}{4}\right)} {x} = \frac{\log(1+t)} {4t}. \] Hence, \[ = \frac14 \left( \frac{\log(1+t)}{t} \right). \] Taking the limit as \(t\to0\), \[ \lim_{t\to0} \frac{\log(1+t)}{t} = 1. \] Therefore, \[ \lim_{x\to0} \frac{\log(4+x)^x-\log4^x} {\sin^2x} = \frac14. \] Thus, \[ \boxed{\frac14}. \]