[venkat jothi]

SORRY LaTeX  DOES NOT WORK  FOR SOME EQUATIONS IN MY ABOVE SOLUTION

[venkat jothi]

by using the definition of permutation we can get an equivalent form

$1+1\times 1!+2\times 2!+3\times 3!+\cdots+n\times n! $

now , $r\times r! =r\times r! +r!-r!= (r+1)r!-r!$

now, $1\times 1!=2!-1!,$

and $2\times 2!=3!-2!$

and so on

$n\times n!=(n+1)!-n! $

the first equation becomes,

$1+2!-1!+3!-2!+4!-3!+\cdots+(n+1)!-n! = (n+1)!$

by converting the terms into permutation definition we get above result

[venkat jothi]

given that f(x) is a differentiable function.

$$lim_{\delta \to 0} \big(\frac{f(x+\delta x)}{f(x)}\big)^{\frac{1}{\delta}} ---------(1)$$

by direct substitution it is in the form of $1^\infty$.

so we can use this formula

$$lim_{\delta \to 0} \big(1+g(\delta)\big)^{h(\delta)}=e^{lim_{\delta \to 0} g(\delta).h(\delta)$$

here , from(1) we get

$$\displaystyle \Rightarrow lim_{\delta \to 0} \big(1+\frac{f(x+\delta x)}{f(x)} -1\big)^{\frac{1}{\delta}}$$

$$\displaystyle \Rightarrow lim_{\delta \to 0} \big(1+\frac{f(x+\delta x)-f(x)}{f(x)}\big)^{\frac{1}{\delta}}$$

here, $g(\delta)=\frac{f(x+\delta x)-f(x)}{f(x)}$ and $h(\delta )= \frac{1}{\delta}$

$$\displaystyle \Rightarrow e^{lim_{\delta \to 0} g(\delta).h(\delta)$$

$$\displaystyle \Rightarrow e^{lim_{\delta \to 0}\frac{f(x+\delta x)-f(x)}{f(x)} .\frac{1}{\delta}$$

by rearranging terms

$$\Rightarrow e^{lim_{\delta \to 0}\frac{f(x+\delta x)-f(x)}{\delta} .\frac{1}{f(x)}$$

by replacing the Definition of derivative $lim_{\delta \to 0}\frac{f(x+\delta x)-f(x)}{\delta} =f'(x)$. we get

$$\Rightarrow e^{\frac{f'(x)}{f(x)}}$$

 

[venkat jothi]

[venkat jothi]

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