If the function f(x) = frac{e^{x}(e^{tan x-x} | vedclass

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(A) For the function to be continuous at $x=0$, $f(0) = \lim_{x 	o 0} f(x)$.$f(0) = \lim_{x 	o 0} \frac{e^{	an x} - e^x + \ln(\sec x + 	an x) - x}

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$2$

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(A) For the function to be continuous at $x=0$, $f(0) = \lim_{x 	o 0} f(x)$.$f(0) = \lim_{x 	o 0} \frac{e^{	an x} - e^x + \ln(\sec x + 	an x) - x}{	an x - x}$.Using the Taylor series expansions near $x=0$:$e^{	an x} = 1 + (x + \frac{x^3}{3}) + \frac{(x + \frac{x^3}{3})^2}{2} + \frac{(x + \frac{x^3}{3})^3}{6} + \dots = 1 + x + \frac{x^2}{2} + \frac{2x^3}{3} + O(x^4)$.$e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + O(x^4)$.$\ln(\sec x + 	an x) = x + \frac{x^3}{6} + O(x^5)$.Substituting these into the numerator:$N = (1 + x + \frac{x^2}{2} + \frac{2x^3}{3}) - (1 + x + \frac{x^2}{2} + \frac{x^3}{6}) + (x + \frac{x^3}{6}) - x = \frac{2x^3}{3} + O(x^4)$.The denominator is $	an x - x = (x + \frac{x^3}{3} + O(x^5)) - x = \frac{x^3}{3} + O(x^5)$.Thus, $f(0) = \lim_{x 	o 0} \frac{\frac{2x^3}{3}}{\frac{x^3}{3}} = \frac{2}{1} = 2$.

If the function $f(x) = \frac{e^{x}(e^{\tan x-x}-1)+\log_{e}(\sec x+\tan x)-x}{\tan x-x}$ is continuous at $x=0$, then the value of $f(0)$ is equal to

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