Question:medium
\( \lim_{x\to0} \frac{1+x-e^x}{x^2} \)
\( \lim_{x\to0} \frac{1+x-e^x}{x^2} \)
Show Hint
Use series expansion when direct substitution gives \(0/0\).
Updated On: May 1, 2026
- \( 1 \)
- \( -\frac{1}{2} \)
- \( 0 \)
- \( \frac{1}{2} \)
- \( -1 \)
Show Solution
The Correct Option is B
Solution and Explanation
Was this answer helpful?
0
Top Questions on limits and derivatives
- Let the function \( g: (-\infty, 0) \rightarrow \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \) be given by \( g(u) = 2 \tan^{-1}(e^u) - \frac{\pi}{2} \). Determine the properties of \( g \).
- BITSAT - 2024
- Mathematics
- limits and derivatives
- Let \( f \) be the function defined by:
\[ f(x) = \begin{cases} \frac{x^2 - 1}{x^2 - 2|x-1| - 1}, & \text{if } x \neq 1, \\ \frac{1}{2}, & \text{if } x = 1. \end{cases} \] The function is continuous at:- BITSAT - 2024
- Mathematics
- limits and derivatives
- Let \( f(x) = x^2 \log(\cos x) \log(1 + x) \) for \( x \neq 0 \), and \( f(0) = 0 \). Determine the behavior of \( f(x) \) at \( x = 0 \).
- BITSAT - 2024
- Mathematics
- limits and derivatives
If \( f(x) \) is defined as follows:
$$ f(x) = \begin{cases} 4, & \text{if } -\infty < x < -\sqrt{5}, \\ x^2 - 1, & \text{if } -\sqrt{5} \leq x \leq \sqrt{5}, \\ 4, & \text{if } \sqrt{5} \leq x < \infty. \end{cases} $$ If \( k \) is the number of points where \( f(x) \) is not differentiable, then \( k - 2 = \)
- BITSAT - 2024
- Mathematics
- limits and derivatives
- Want to practice more? Try solving extra ecology questions todayView All Questions
