Question:
\(\displaystyle\lim_{x\rightarrow0}\frac{\log(1+x)+1-e^x}{4x^2-9x} \ is\ equal\ to\)
\(\displaystyle\lim_{x\rightarrow0}\frac{\log(1+x)+1-e^x}{4x^2-9x} \ is\ equal\ to\)
Updated On: May 31, 2024
- \(\frac{-1}{9}\)
- \(\frac{1}{9}\)
- \(\frac{-1}{18}\)
- \(\frac{1}{18}\)
- 0
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