Question

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

Answer 1

Step-by-step explanation:

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see this attachment

Answer 2

Question:-

$\rm \displaystyle \: \lim_{x \: \to \: 0} \: \rm\frac{(1 + x) {}^{ \frac{1}{x} } - e + \frac{ex}{2} }{ {x}^{2} }$

Solution:-

$\rm \displaystyle \: \lim_{x \: \to \: 0} \: \rm\frac{(1 + x) {}^{ \frac{1}{x} } - e + \frac{ex}{2} }{ {x}^{2} }$

To solve this question first we know the expansion of

$\rm \: (1 + x) {}^{ \frac{1}{x} } = e(1 - \dfrac{x}{2} + \dfrac{11}{24} {x}^{2} - \: \: .........)$

But we have to need only

$\rm \: (1 + x) {}^{ \frac{1}{x} } = e(1 - \dfrac{x}{2} + \dfrac{11}{24} {x}^{2} )$

Now put the value

$\rm \displaystyle \: \lim_{x \: \to \: 0} \: \rm\frac{e(1 - \dfrac{x}{2} + \dfrac{11}{24} {x}^{2} ) {}^{ } - e + \frac{ex}{2} }{ {x}^{2} }$

$\rm \displaystyle \: \lim_{x \: \to \: 0} \: \rm\frac{(e- \dfrac{xe}{2} + \dfrac{11}{24} {x}^{2} ) {}^{ } - e + \frac{ex}{2} }{ {x}^{2} }$

$\rm \displaystyle \: \lim_{x \: \to \: 0} \: \rm\frac{ \cancel e- \cancel \dfrac {xe}{2} + \dfrac{11e}{24} \cancel {x}^{2} {}^{ } - \cancel e + \cancel \dfrac{ex}{2} }{ \cancel {x}^{2} }$

Now we get

$\rm \displaystyle \: \lim_{x \: \to \: 0} \: \rm \frac{11e}{24}$

Answer:-

$\boxed{ \rm = \frac{11e}{24} }$