Question

Don't use binomial theorem concept.

[tex]\displaystyle \lim_{x \to 0} \frac{(x+1)^5-1}{x-2}

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Answer 1

${\huge{\underbrace{\rm{Answer\checkmark}}}}$

$\displaystyle \lim_{x \to 0} \frac{(x+1)^5-1}{x-2}$

$\implies \: \dfrac{{(0 + 1)}^{5} - 1 }{0 - 2}$

$\implies \: \dfrac{1 - 1}{ - 2}$

$\implies \: 0$

$\rm \: putting \: y = x + 1$

$\longrightarrow \rm \: x = y - 1$

$\rm as \: \: \: \: x\to0$

$\rm \: \: \: y \to \: 0 + 1$

$\rm \: \: \: y \to1$

Now , new equation would be

$\displaystyle\lim_{y \to 1} \frac{ {y}^{5} - 1 }{y - 3}$

I think Question is wrong it , must be

$\displaystyle \lim_{x \to 0} \frac{(x+1)^5-1}{x}$

Answer 2

Step-by-step explanation:

Newton's second law states that the acceleration of an object is directly related to the net force and inversely related to its mass. Acceleration of an object depends on two things, force and mass.