Question

$\lim_{x \to 0} \dfrac{\left(1+x\right)^{\frac{1}{2}}-e+\dfrac{ex}{2}}{x^2}$ equals

Answer 1

Some related formula of expansion.

$\bold{1) = log(1 + x) = x - \frac{x {}^{2} }{2} + \frac{x {}^{3} }{3} - \frac{ {x}^{4} }{4}... }$

$\bold{2) = log(1 - x) = - x - \frac{ {x}^{2} }{2} - \frac{ {x}^{3} }{3} - \frac{ {x}^{4} }{4} ..... }$

$\bold{3) = e {}^{x} = 1 + x + \frac{ {x}^{2} }{ 2!} + \frac{ {x}^{3} }{3!} + ...}$

$\bold{4) = e {}^{ - x} = 1 - x + \frac{ {x}^{2} }{2!} - \frac{ {x}^{3} }{3!} + ... }$

$\bold{5) = \sin(x) = x - \frac{ {x}^{3} }{3!} + \frac{ {x}^{5} }{5!} - \frac{ {x}^{7} }{7!} ...... }$

$\bold{6) = \cos(x) = 1 - \frac{ {x}^{2} }{2!} + \frac{ {x}^{4} }{4!} - \frac{ {x}^{6} }{6!} ....}$

$\bold{7) = \tan(x) = x + \frac{ {x}^{3} }{3} + \frac{2 {x}^{5} }{15} + ....}$

Note = for solution see image

attachment.

Answer 2

Step-by-step explanation:

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see this attachment