If $\displaystyle\lim_{x \to 5} \frac{x^k - 5^k}{x - 5} = 500$ then k is equal to :

Let $\displaystyle\lim_{x \to 5} \frac{x^k - 5^k}{x - 5} = 500$ By using $\displaystyle\lim_{x \to a} \frac{x^n - a^n}{x - a} = n.a^{n -1} $ we have $k.5^{k -1} = 500$ Now, put $k = 4$ , we get $4.5^{4 -1} = 500$ $\Rightarrow \:\: 4.5^3 = 500$ which is ture. $\therefore\:\:\: k = 4$

If $\displaystyle\lim_{x \to 5} \frac{x^k

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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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 = $
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Notes on limits and derivatives

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:

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