Question

plz answer this ...​

Answer 1

Answer:

Use

$for \: \: x \to0 \: \: {(1 + x)}^{k} \approx \: kx$

So now your limit

$\lim_{x \to0}( \frac{1 + kx - 1}{x} ) = k$

You can also use L hopital rule or full series expansion , choice is yours