$\displaystyle \lim_{x \to 1}$$\left[\left(\frac{4x}{x^{2}-x^{-1}}-\frac{1-3x+x^{2}}{1-x^{3}}\right)^{-1}+3\left(\frac{x^{4}-1}{x^{3}-x^{-1}}\right)\right]$ is
- $2$
- $3$
- $4$
- $1$
The Correct Option is B
Solution and Explanation
Top Questions on limits and derivatives
Let $\left\lfloor t \right\rfloor$ be the greatest integer less than or equal to $t$. Then the least value of $p \in \mathbb{N}$ for which
\[ \lim_{x \to 0^+} \left( x \left\lfloor \frac{1}{x} \right\rfloor + \left\lfloor \frac{2}{x} \right\rfloor + \dots + \left\lfloor \frac{p}{x} \right\rfloor \right) - x^2 \left( \left\lfloor \frac{1}{x^2} \right\rfloor + \left\lfloor \frac{2}{x^2} \right\rfloor + \dots + \left\lfloor \frac{9^2}{x^2} \right\rfloor \right) \geq 1 \]
is equal to __________.
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- 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
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- If \( f(x) \) is defined as follows:
\[ f(x) = \begin{cases} 4, & {if } -\infty < x < -\sqrt{5}, \\ x^2 - 1, & {if } -\sqrt{5} \leq x \leq \sqrt{5}, \\ 4, & {if } \sqrt{5} \leq x < \infty. \end{cases} \] If \( k \) is the number of points where \( f(x) \) is not differentiable, then \( k - 2 = \)- BITSAT - 2024
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Notes on limits and derivatives
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Concepts Used:
Limits And Derivatives
Mathematically, a limit is explained as a value that a function approaches as the input, and it produces some value. Limits are essential in calculus and mathematical analysis and are used to define derivatives, integrals, and continuity.
Limits Formula:
Derivatives of a Function:
A derivative is referred to the instantaneous rate of change of a quantity with response to the other. It helps to look into the moment-by-moment nature of an amount. The derivative of a function is shown in the below-given formula.
Properties of Derivatives:
Read More: Limits and Derivatives