Question

\(\lim_{x\to0}\dfrac{ln(1+(ln5)x)}{5^{x}}-1\)

• A. \(1\)
• B. \(ln 5\)
• C. \(-1\)
• D. \(5\)
• E. \(\dfrac{1}{5}\)

Answer 1

The Correct Option is C

 We want to evaluate:

\[ \lim_{x \to 0} \left[ \frac{\ln(1 + (\ln 5)x)}{5^x} - 1 \right] \]

This can be rewritten as:

\[ \lim_{x \to 0} \left[ \frac{\ln(1 + (\ln 5)x)}{5^x} \right] - \lim_{x \to 0} 1 \]

The second term is simply \(-1\).

Let's focus on the first term:

\[ \lim_{x \to 0} \left[ \frac{\ln(1 + (\ln 5)x)}{5^x} \right] \]

This limit is of the indeterminate form \( \frac{0}{1} \). We can use L'Hopital's rule, but it's easier to use the fact that \( \lim_{u \to 0} \frac{\ln(1+u)}{u} = 1 \).

Let \( u = (\ln 5)x \). Then as \( x \to 0 \), \( u \to 0 \). Also, note that \( 5^x \to 1 \) as \( x \to 0 \).

\[ \lim_{x \to 0} \left[ \frac{\ln(1 + (\ln 5)x)}{5^x} \right] = \lim_{u \to 0} \left[ \frac{\ln(1 + u)}{5^{u / \ln 5}} \right] = \lim_{u \to 0} \frac{\ln(1 + u)}{u} \cdot \lim_{u \to 0} \frac{u}{5^{u / \ln 5}} \]

Since \( \lim_{u \to 0} \frac{\ln(1 + u)}{u} = 1 \) and \( \lim_{u \to 0} \frac{u}{5^{u / \ln 5}} = \frac{0}{1} = 0 \), then the entire limit will be \( 0 \).

Therefore:

\[ \lim_{x \to 0} \left[ \frac{\ln(1 + (\ln 5)x)}{5^x} - 1 \right] = 0 - 1 = -1 \]

Answer 2

Let's evaluate the following limit:

\[ \lim_{x \to 0} \left( \frac{\ln(1 + (\ln 5) x)}{5^x} - 1 \right) \]

Step 1: Expand the logarithmic function

The logarithmic function \( \ln(1 + (\ln 5) x) \) can be expanded using the Taylor series for \( \ln(1 + y) \) around \( y = 0 \). The series expansion is:

\[ \ln(1 + y) = y - \frac{y^2}{2} + O(y^3) \]

Substituting \( y = (\ln 5) x \), we get:

\[ \ln(1 + (\ln 5) x) = (\ln 5) x - \frac{(\ln 5)^2 x^2}{2} + O(x^3) \]

Step 2: Expand \( 5^x \)

Next, we need the expansion of \( 5^x \) as \( x \to 0 \). Using the series expansion for \( a^x \) around \( x = 0 \), we have:

\[ 5^x = e^{x \ln 5} = 1 + x \ln 5 + \frac{x^2 (\ln 5)^2}{2} + O(x^3) \]

Step 3: Plug the expansions into the limit expression

Now, substitute the expansions into the original limit expression:

\[ \frac{\ln(1 + (\ln 5) x)}{5^x} = \frac{(\ln 5) x - \frac{(\ln 5)^2 x^2}{2} + O(x^3)}{1 + x \ln 5 + \frac{x^2 (\ln 5)^2}{2} + O(x^3)} \]

We can now approximate this expression for small \( x \). For \( x \to 0 \), the terms involving higher powers of \( x \) (like \( x^2 \) and higher) can be ignored for the leading order approximation. Thus:

\[ \frac{\ln(1 + (\ln 5) x)}{5^x} \approx \frac{(\ln 5) x}{1 + x \ln 5} \approx (\ln 5) x \quad \text{(as higher terms vanish for small \( x \))} \]

Step 4: Subtract 1 and compute the limit

Now, we subtract 1 from the expression:

\[ \frac{\ln(1 + (\ln 5) x)}{5^x} - 1 \approx (\ln 5) x - 1 \]

Finally, take the limit as \( x \to 0 \):

\[ \lim_{x \to 0} \left( (\ln 5) x - 1 \right) = -1 \]