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# Sets1

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• #66965
Shreya Nair
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Sets1

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#71214
venkat jothi
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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

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

• This reply was modified 3 months, 4 weeks ago by venkat jothi.
#71261
venkat jothi
Participant

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

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