I don't think limits - U. Bolt

It is in (0/0) form so we can use L Hospital's rule.But it will be more easier if you use the expansion

${ \left( 1+x \right) }^{ \cfrac { 1 }{ x } }=e-\cfrac { ex }{ 2 } +\cfrac { 11e{ x }^{ 2 } }{ 24 } -\cfrac { 7e{ x }^{ 3 } }{ 16 } +...\\ \\ \Rightarrow y=e-{ \left( 1+x \right) }^{ \cfrac { 1 }{ x } }=\cfrac { ex }{ 2 } -\cfrac { 11e{ x }^{ 2 } }{ 24 } +\cfrac { 7e{ x }^{ 3 } }{ 16 } -...\\ \\ \Rightarrow \cfrac { dy }{ dx } =\cfrac { e }{ 2 } -\cfrac { 11e{ x } }{ 12 } +\cfrac { 21e{ x }^{ 2 } }{ 16 } -...\\ \\ \Rightarrow \underset { x\rightarrow 0 }{ lim } \cfrac { e-{ \left( 1+x \right) }^{ \cfrac { 1 }{ x } } }{ \tan { x } } \\ \\ \quad =\underset { x\rightarrow 0 }{ lim } \cfrac { \cfrac { e }{ 2 } -\cfrac { 11e{ x } }{ 12 } +\cfrac { 21e{ x }^{ 2 } }{ 16 } -... }{ \sec ^{ 2 }{ x } } \quad \left( L\quad Hospital'srule \right) \\ \\ \quad =\cfrac { e }{ 2 } =1.359$

I am adding proof for used expansion,

$f\left( x \right) =y={ \left( 1+x \right) }^{ 1/x }\Rightarrow \underset { x\rightarrow 0 }{ lim } { y }=e\\ \\ \ln { y } =\cfrac { \ln { \left( 1+x \right) } }{ x } =1-\cfrac { x }{ 2 } +\cfrac { { x }^{ 2 } }{ 3 } -\cfrac { { x }^{ 3 } }{ 4 } +\cfrac { { x }^{ 4 } }{ 5 } -...\\ \\ \Rightarrow { y }_{ 1 }=y\left( \cfrac { -1 }{ 2 } +\cfrac { 2x }{ 3 } -\cfrac { 3{ x }^{ 2 } }{ 4 } +\cfrac { 4{ x }^{ 3 } }{ 5 } -... \right) \\ \\ \Rightarrow \underset { x\rightarrow 0 }{ lim } { y }_{ 1 }=\underset { x\rightarrow 0 }{ lim } { y }\left( \cfrac { -1 }{ 2 } +\cfrac { 2x }{ 3 } -\cfrac { 3{ x }^{ 2 } }{ 4 } +\cfrac { 4{ x }^{ 3 } }{ 5 } -... \right) =\cfrac { -e }{ 2 } \\ \\ \\ Now,{ y }_{ 2 }={ y }_{ 1 }\left( \cfrac { -1 }{ 2 } +\cfrac { 2x }{ 3 } -\cfrac { 3{ x }^{ 2 } }{ 4 } +\cfrac { 4{ x }^{ 3 } }{ 5 } -... \right) \\ \\ +y\left( \cfrac { 2 }{ 3 } -\cfrac { 6x }{ 4 } +\cfrac { 12{ x }^{ 2 } }{ 5 } -... \right) \\ \\ \Rightarrow \underset { x\rightarrow 0 }{ lim } { y }_{ 2 }=\underset { x\rightarrow 0 }{ lim } { y }_{ 1 }\left( \cfrac { -1 }{ 2 } +\cfrac { 2x }{ 3 } -\cfrac { 3{ x }^{ 2 } }{ 4 } +\cfrac { 4{ x }^{ 3 } }{ 5 } -... \right) \\ \\ +\underset { x\rightarrow 0 }{ lim } y\left( \cfrac { 2 }{ 3 } -\cfrac { 6x }{ 4 } +\cfrac { 12{ x }^{ 2 } }{ 5 } -... \right) \\ \\ =\left( \cfrac { -e }{ 2 } \right) \left( \cfrac { -1 }{ 2 } \right) +e\left( \cfrac { 2 }{ 3 } \right) =\cfrac { 11e }{ 12 } \\ \\ \\ \& \quad { y }_{ 3 }={ y }_{ 2 }\left( \cfrac { -1 }{ 2 } +\cfrac { 2x }{ 3 } -\cfrac { 3{ x }^{ 2 } }{ 4 } +\cfrac { 4{ x }^{ 3 } }{ 5 } -... \right) \\ \\ +{ y }_{ 1 }\left( \cfrac { 2 }{ 3 } -\cfrac { 6x }{ 4 } +\cfrac { 12{ x }^{ 2 } }{ 5 } -... \right) \\ \\ \quad +{ y }_{ 1 }\left( \cfrac { 2 }{ 3 } -\cfrac { 6x }{ 4 } +\cfrac { 12{ x }^{ 2 } }{ 5 } -... \right) +y\left( \cfrac { -6 }{ 4 } +\cfrac { 24x }{ 5 } -... \right) \\ \\ ={ y }_{ 2 }\left( \cfrac { -1 }{ 2 } +\cfrac { 2x }{ 3 } -\cfrac { 3{ x }^{ 2 } }{ 4 } +\cfrac { 4{ x }^{ 3 } }{ 5 } -... \right) \\ \\ +{ 2y }_{ 1 }\left( \cfrac { 2 }{ 3 } -\cfrac { 6x }{ 4 } +\cfrac { 12{ x }^{ 2 } }{ 5 } -... \right) +y\left( \cfrac { -6 }{ 4 } +\cfrac { 24x }{ 5 } -... \right) \\ \\ \\ \Rightarrow \underset { x\rightarrow 0 }{ lim } { y }_{ 3 }=\underset { x\rightarrow 0 }{ lim } { y }_{ 2 }\left( \cfrac { -1 }{ 2 } +\cfrac { 2x }{ 3 } -\cfrac { 3{ x }^{ 2 } }{ 4 } +\cfrac { 4{ x }^{ 3 } }{ 5 } -... \right) \\ \\ +\underset { x\rightarrow 0 }{ lim } 2{ y }_{ 1 }\left( \cfrac { 2 }{ 3 } -\cfrac { 6x }{ 4 } +\cfrac { 12{ x }^{ 2 } }{ 5 } -... \right) +\underset { x\rightarrow 0 }{ lim } y\left( \cfrac { -6 }{ 4 } +\cfrac { 24x }{ 5 } -... \right) \\ \\ =\left( \cfrac { 11e }{ 12 } \right) \left( \cfrac { -1 }{ 2 } \right) +2\left( \cfrac { -e }{ 2 } \right) \left( \cfrac { 2 }{ 3 } \right) +e\left( \cfrac { -6 }{ 4 } \right) =\cfrac { -21e }{ 8 } \\ \\ or\quad \underset { x\rightarrow 0 }{ lim } { f\left( x \right) }=e,\underset { x\rightarrow 0 }{ lim } { f^{ ' }\left( x \right) }=\cfrac { -e }{ 2 } ,\underset { x\rightarrow 0 }{ lim } { f^{ '' }\left( x \right) }=\cfrac { 11e }{ 12 } \\ \\ \& \underset { x\rightarrow 0 }{ lim } { f^{ ''' }\left( x \right) }=\cfrac { -21e }{ 8 } \\ \\ \\ for\quad Taylor's\quad series\quad expansion\quad at\quad x=0,\\ \\ f\left( x \right) =f\left( 0 \right) +\cfrac { { f^{ ' }\left( 0 \right) } }{ 1! } +\cfrac { { f^{ '' }\left( 0 \right) } }{ 2! } +\cfrac { { f^{ ''' }\left( 0 \right) } }{ 3! } +...\\ \\ \Rightarrow y={ \left( 1+x \right) }^{ 1/x }=e-\cfrac { ex }{ 2 } +\cfrac { 11e{ x }^{ 2 } }{ 24 } -\cfrac { 7e{ x }^{ 3 } }{ 16 } +...$

$\large \lim _{ x\to 0 } \frac { e-(1+x)^{ \frac { 1 }{ x } } }{ \tan x }$ $\large \lim _{ x\to 0 } \frac { e-{ e }^{ \frac { \ln { (1+x) } }{ x } } }{ \tan x }$ $\large \lim _{ x\to 0 } \frac { -e({ e }^{ \frac { \ln { (1+x) } }{ x } -1 }-1) }{ \tan x }$ $\large\lim _{ x\to 0 } \frac { -e({ e }^{ \frac { \ln { (1+x) } }{ x } -1 }-1)(\frac { \ln { (1+x) } }{ x } -1)x }{ \tan x(\frac { \ln { (1+x) } }{ x } -1)x }$ $(\lim _{ x\to 0 } -e)(\lim _{ x\to 0 } \frac { ({ e }^{ \frac { \ln { (1+x) } }{ x } -1 }-1) }{ (\frac { \ln { (1+x) } }{ x } -1) } )(\lim _{ x\to 0 } \frac { x }{ \tan x } )(\lim _{ x\to 0 } \frac { \ln { (1+x) } -{ x } }{ { x }^{ 2 } } )$ $\large(-e)(\lim _{ x\to 0 } \frac { \ln { (1+x) } -{ x } }{ { x }^{ 2 } } )$ $\large(-e)(\lim _{ x\to 0 } \frac { -\frac { { x }^{ 2 } }{ 2 } +\frac { { x }^{ 3 } }{ 3 } -\frac { { x }^{ 4 } }{ 4 } +\dots }{ { x }^{ 2 } } )$ $\large(-e)(\lim _{ x\to 0 } \frac { -1 }{ 2 } +\frac { x }{ 3 } -\frac { { x }^{ 2 } }{ 4 } +\dots )$ $\large\boxed { \frac { e }{ 2 } }$

First let's get rid of that $\tan(x)$

$\displaystyle L= \lim_{x \rightarrow 0} \frac{\frac{e-(1+x)^{1/x}}{x}}{\frac{\tan x}{x}}=\lim_{x \rightarrow 0} \frac{e-(1+x)^{1/x}}{x}$

Method 1 (L'hopital's rule)

The function is in $\frac{0}{0}$ form. So we can use l'Hopital's rule. Find the derivative of $(1+x)^{1/x}$ using logarithmic differentiation.

$\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} (1+x)^{1/x} = (1+x)^{1/x}(\frac{x-(1+x)\ln(1+x)}{x^2(1+x)})$

and $\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} x = 1$

Therefore, from l'Hopital's rule

$\displaystyle L=\lim_{x \rightarrow 0} -(1+x)^{1/x}(\frac{x-(1+x)\ln(1+x)}{x^2(1+x)}) = -e \lim_{x \rightarrow 0} \frac{x-(1+x)\ln(1+x)}{x^2(1+x)}$

Use l'Hopital's rule again (because 0/0 form)

$\displaystyle L=-e \lim_{x \rightarrow 0} \frac{-\ln(1+x)}{2x(1+x) + x^2} = -e (\frac{-1}{2}) = \boxed{\frac{e}{2}}$

Method 2 (Maclaurin Series)

You may remember for $x \rightarrow 0$

$\displaystyle (1+x)^{1/x} = e - \frac{ex}{2} + \frac{11ex^2}{24} - O(x^3)$

$\displaystyle e - (1+x)^{1/x} = \frac{ex}{2} - \frac{11ex^2}{24} +O(x^3)$

$\displaystyle \frac{e - (1+x)^{1/x}}{x} = \frac{e}{2} - \frac{11ex}{24} +O(x^2)$

And finally $\displaystyle \lim_{x \rightarrow 0} \frac{e - (1+x)^{1/x}}{x} = \boxed{\frac{e}{2}}$

If you don't remember the maclaurin series, it's usually better to use l'Hopital's rule since you'd have to find the derivatives anyway. But if you insist you can still find the first two terms of the maclaurin series (that's all we need to solve the problem) without the derivatives in the following manner.

In a sufficiently small neighbourhood of 0 (since $x \rightarrow 0$ ), we can use the binomial expansion

$\displaystyle (1+x)^{1/x} = 1+\frac{1}{1!} + \frac{1(1-x)}{2!} + \frac{1(1-x)(1-2x)}{3!} + \frac{1(1-x)(1-2x)(1-3x)}{4!} +\dots$

$\displaystyle =(1+\frac{1}{1!} +\frac{1}{2!}+\frac{1}{3!} +\dots)-(\frac{1}{2!}+\frac{1}{3!}+\frac{2}{3!}+\frac{1}{4!}+\frac{2}{4!}+\frac{3}{4!}+\frac{1}{5!}+\dots)x +O(x^2)$

$\displaystyle = e - (\sum_{r=2}^{\infty} \frac{\sum_{k=1}^{r-1} k}{r!})x +O(x^2)$

$\displaystyle = e - (\sum_{r=2}^{\infty} \frac{(r-1)(r)}{2(r!)})x + O(x^2)$

$\displaystyle = e - (\frac{1}{2} \sum_{r=2}^{\infty} \frac{1}{(r-2)!})x + O(x^2)$

$\displaystyle = e -\frac{ex}{2} +O(x^2)$

And you can solve the problem as shown above.

$L =\lim_{x\to 0} \dfrac{e - \,(1+x)^{\frac{1}{x}}}{\tan x} \\ L = \lim_{x\to 0} \dfrac{e -\left(e -\dfrac{ex}{2} +\dfrac{11ex^2}{12}+O(x^3)\right) }{ x +\dfrac{x^3}{3} +\dfrac{2x^5}{5}+\cdots } \\ L = \lim_{x\to 0} \dfrac{xe\left(\dfrac{1}{2} -\dfrac{11x}{12}+O(x^3)\right) }{ x\left(1 +\dfrac{x^2}{3} +\dfrac{2x^4}{5}+\cdots \right)} \\ L = \dfrac{e}{2} = \frac{1}{2}\times e^1$ Hence $A+B = 1+\frac{1}{2}= \frac{3}{2}=\boxed{1.5}$