Question

\(\displaystyle\lim_{x\rightarrow0}\frac{\log(1+x)+1-e^x}{4x^2-9x} \ is\ equal\ to\)

• A. \(\frac{-1}{9}\)
• B. \(\frac{1}{9}\)
• C. \(\frac{-1}{18}\)
• D. \(\frac{1}{18}\)
• E. 0

Answer 1

Answer: (E)

We are tasked with finding the limit: \[ \lim_{x \to 0} \frac{\log(1 + x) + 1 - e^x}{4x^2 - 9x} \]
Step 1: Apply Taylor series expansions for \( \log(1+x) \) and \( e^x \)
The Taylor series expansion of \( \log(1+x) \) around \( x = 0 \) is: \[ \log(1+x) = x - \frac{x^2}{2} + O(x^3) \] The Taylor series expansion of \( e^x \) around \( x = 0 \) is: \[ e^x = 1 + x + \frac{x^2}{2} + O(x^3) \] Thus, we can approximate: \[ \log(1 + x) + 1 - e^x = \left( x - \frac{x^2}{2} + O(x^3) \right) + 1 - \left( 1 + x + \frac{x^2}{2} + O(x^3) \right) \] Simplifying: \[ = x - \frac{x^2}{2} + 1 - 1 - x - \frac{x^2}{2} + O(x^3) \] \[ = -x^2 + O(x^3) \] Step 2: Simplify the denominator The denominator is \( 4x^2 - 9x \), which is already in a simplified form.
Step 3: Substitute the approximations into the original expression Substituting \( \log(1 + x) + 1 - e^x = -x^2 + O(x^3) \) into the expression: \[ \frac{-x^2 + O(x^3)}{4x^2 - 9x} \] For small \( x \), the higher-order terms \( O(x^3) \) can be neglected. Therefore, the expression simplifies to: \[ \frac{-x^2}{4x^2 - 9x} \] Factor out \( x \) from the denominator: \[ = \frac{-x^2}{x(4x - 9)} \] \[ = \frac{-x}{4x - 9} \] ### Step 4: Take the limit as \( x \to 0 \) As \( x \to 0 \), the expression simplifies to: \[ \lim_{x \to 0} \frac{-x}{4x - 9} = \frac{0}{-9} = 0 \]

The correct option is (E) : \(0\)

Answer 2

We want to evaluate the limit: \[\displaystyle\lim_{x\rightarrow0}\frac{\log(1+x)+1-e^x}{4x^2-9x}\]

1. We can use L'Hôpital's Rule since the limit is of the form \(\frac{0}{0}\). Let's find the derivatives of the numerator and denominator.
2. The derivative of the numerator is: \[\frac{d}{dx} (\log(1+x)+1-e^x) = \frac{1}{1+x} - e^x \]
3. The derivative of the denominator is: \[\frac{d}{dx} (4x^2-9x) = 8x-9 \]
4. Now we have: \[\displaystyle\lim_{x\rightarrow0}\frac{\frac{1}{1+x} - e^x}{8x-9}\]
5. Plugging in \(x=0\), we get: \[\frac{\frac{1}{1+0} - e^0}{8(0)-9} = \frac{1-1}{-9} = \frac{0}{-9} = 0\]
6. Therefore, the original limit is 0. However, since the initial answer choices don't contain 0, we need to consider higher order terms to resolve the \(\frac{0}{0}\) form to the first order. Let's use Taylor series expansions.
7. Taylor Expansion: \(\log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - ...\) \(e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + ...\)
8. Substitute Taylor Series into the numerator: \(\log(1+x) + 1 - e^x = (x - \frac{x^2}{2} + ...) + 1 - (1 + x + \frac{x^2}{2} + ...) = -x^2 + O(x^3)\)
9. So the limit becomes: \(\displaystyle\lim_{x\rightarrow0}\frac{-x^2}{4x^2-9x} = \displaystyle\lim_{x\rightarrow0}\frac{-x}{4x-9}\)
10. Now, we can directly evaluate the limit by plugging in x=0: \(\frac{-0}{4(0)-9} = \frac{0}{-9} = 0\)
11. Since the answer choice includes \(\frac{-1}{18}\), \(\frac{-1}{9}\), \(\frac{1}{18}\) and \(\frac{1}{9}\), let's consider one more differentiation using L'Hopital's rule on step 5. \[ \lim_{x \to 0} \frac{\frac{1}{1+x} - e^x}{8x-9} \] Applying L'Hopital's rule will result in increasingly complex derivatives. Instead, let us substitute the series from step 8 into the original limit expression and solve it. \[ \lim_{x \to 0} \frac{(x - \frac{x^2}{2} + O(x^3)) + 1 - (1 + x + \frac{x^2}{2} + O(x^3))}{4x^2-9x} = \lim_{x \to 0} \frac{-x^2 + O(x^3)}{-9x + 4x^2} \] Dividing top and bottom by x, gives \[\lim_{x \to 0} \frac{-x + O(x^2)}{-9 + 4x} = \frac{0}{-9} = 0 \]
12. Since the answer is still 0, and since 0 is a valid choice in the given answers, the final answer is 0.