Question

limit X tends to zero e raised to X minus 1 whole divided by X is one how​

Answer 1

$\huge{ \underline{ \underline{ \green{ \sf{ Detailed \: Answer :}}}}}$

${\red{\huge{\underline{\mathbb{To \: Prove :-}}}}}$

$\star \bold{ \lim_{ \Delta x \to \: 0}( \frac{e{}^{x} - 1}{ x} )} = 1$

${\orange {\huge{\underline{\mathbb{Answer:-}}}}}$

We know the expansion of $e^x$

${\blue{\boxed{\large{\bold{e{}^{x}= 1+x + \frac{ x{ }^{2} } {2 !} + \frac{x {}^{3} }{3!} +\frac{ x{}^{4} }{4!} .....}}}}}$

$\star \bold{ \lim_{ \Delta x \to \: 0}( \frac{e{}^{x} - 1}{ x} )}$

Using expansion of $e^x$

$\star \bold{ \lim_{ \Delta x \to \: 0}( \frac{(1 + x + \frac{x {}^{2} }{2!} + \frac{x {}^{3} }{3!} + \frac{x {}^{4} }{4!} .... ) - 1}{ x} )}$

$\star \bold{ \lim_{ \Delta x \to \: 0}( \frac{(1 +( x + \frac{x {}^{2} }{2!} + \frac{x {}^{3} }{3!} + \frac{x {}^{4} }{4!} .... ) - 1}{x} )}$

$\star \bold{ \lim_{ \Delta x \to \: 0}( \frac{( x + \frac{x {}^{2} }{2!} + \frac{x {}^{3} }{3!} + \frac{x {}^{4} }{4!} .... )}{ x} )}$

$\star \bold{ \lim_{ \Delta x \to \: 0}( \frac{x +( 1 + \frac{x {}^{} }{2!} + \frac{x {}^{2} }{3!} + \frac{x {}^{3} }{4!} .... )}{ x} )}$

$\star \bold{ \lim_{ \Delta x \to \: 0}( \frac{x }{ x} )}$

= 1

$\huge{\bold{ Hence \: Proved }}$