Question

Evaluate : $\lim_{x\to 0} \left[ \frac{\log 4 + \log(0.25+x)}{x} \right]$

Answer 1

4

Answer 2

The problem asks to evaluate the limit: $\lim_{x\to 0} \left[ \frac{\log 4 + \log(0.25+x)}{x} \right]$

We assume $\log$ denotes the natural logarithm ($\ln$), which is standard in calculus unless specified otherwise.

Step 1: Simplify the numerator using logarithm properties. The numerator is $\log 4 + \log(0.25+x)$. Using the logarithm property $\log A + \log B = \log (AB)$: $\log 4 + \log(0.25+x) = \log \left( 4 \times (0.25+x) \right)$ $= \log \left( 4 \times \frac{1}{4} + 4x \right)$ $= \log (1 + 4x)$

Step 2: Rewrite the limit expression. Substitute the simplified numerator back into the limit: $\lim_{x\to 0} \left[ \frac{\log (1 + 4x)}{x} \right]$

Step 3: Evaluate the limit. As $x \to 0$, the numerator $\log(1+4x) \to \log(1) = 0$. As $x \to 0$, the denominator $x \to 0$. This is an indeterminate form of type $\frac{0}{0}$. We can use either the standard limit formula or L'Hopital's Rule.

Method 1: Using the Standard Limit Formula We know the standard limit formula: $\lim_{y\to 0} \frac{\log (1+y)}{y} = 1$. To apply this, we need the term in the denominator to match the argument of the logarithm. In our case, the argument is $4x$. So, we multiply and divide the expression by 4: $\lim_{x\to 0} \left[ \frac{\log (1 + 4x)}{x} \right] = \lim_{x\to 0} \left[ \frac{\log (1 + 4x)}{4x} \times 4 \right]$ Let $y = 4x$. As $x \to 0$, $y \to 0$. The limit becomes: $4 \times \lim_{y\to 0} \frac{\log (1+y)}{y} = 4 \times 1 = 4$

Method 2: Using L'Hopital's Rule Since the limit is of the $\frac{0}{0}$ form, we can apply L'Hopital's Rule. Let $f(x) = \log(1+4x)$ and $g(x) = x$. Find the derivatives of $f(x)$ and $g(x)$: $f'(x) = \frac{d}{dx} (\log(1+4x)) = \frac{1}{1+4x} \times \frac{d}{dx}(1+4x) = \frac{4}{1+4x}$ $g'(x) = \frac{d}{dx} (x) = 1$ Apply L'Hopital's Rule: $\lim_{x\to 0} \left[ \frac{f'(x)}{g'(x)} \right] = \lim_{x\to 0} \left[ \frac{\frac{4}{1+4x}}{1} \right]$ $= \lim_{x\to 0} \left[ \frac{4}{1+4x} \right]$ Now, substitute $x=0$: $= \frac{4}{1+4(0)} = \frac{4}{1} = 4$

Both methods yield the same result.

The expression is simplified using logarithm properties to $\frac{\log(1+4x)}{x}$. This is a standard limit form $\lim_{y\to 0} \frac{\log(1+y)}{y}=1$. By multiplying and dividing by 4, the limit is transformed to $4 \times \lim_{4x\to 0} \frac{\log(1+4x)}{4x}$, which evaluates to $4 \times 1 = 4$. Alternatively, L'Hopital's Rule can be applied to the $\frac{0}{0}$ form, yielding the same result.