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{\text { If } f(x)=\frac{e^{\tan x}-e^x+\ln (\sec x+\tan x)-x}{\tan x-x} \text { is a continuous function at } x=0 \text {, find } f(0)}

Option: 1

3/2


Option: 2

1


Option: 3

2


Option: 4

0


Answers (1)

best_answer

{\text { If the function } f(x) \text { is continuous at } x=0 \text {, then } f(0)=L=\lim _{x \mid r o 0} f(x) \text {. }}

{\begin{gathered} f(x)=\frac{e^{\tan x}-e^x+\ln (\sec x+\tan x)-x}{\tan x-x} \\ L=\lim _{x \rightarrow 0} \lim _{x \rightarrow 0} \frac{e^{\tan x}-e^x}{\tan x-x}+\lim _{x \rightarrow 0} \frac{\ln (\sec x+\tan x)-x}{\tan x-x} \end{gathered}}

 

{\text { Using } \lim _{z \rightarrow 0} \frac{e^2-1}{z}=1}

                                 {\Longrightarrow 1+\lim _{x \rightarrow 0} \frac{\ln (\tan (x / 2+\pi / 4)-x}{\tan x-x}}

{\text { Use } \ln (1+z)=z-z^2 / 2+z^3 / 3+\ldots, \text { then }}

                                 {\Longrightarrow L=1+\lim _{x \rightarrow 0} \frac{\ln (1+\tan (x / 2)-\ln (1-\tan (x / 2))-x}{\tan x-x}}

{\text { Use } \ln (1+z)=z-z^2 / 2+z^3 / 3+\ldots \text {, then }}

                                 \Longrightarrow L=1+\lim _{x \rightarrow+\ldots 0} \frac{2\left(\tan (x / 2)+(1 / 3) \tan ^3(x / 2)+\ldots\right)-x}{\tan x-x}

{\text { Next, using } \tan z=z+z^3 / 3+\ldots \text {, when } z \text { is very small, wr get }}

                                 {\Longrightarrow L=1+\lim _{x \rightarrow+\ldots 0} \frac{2\left(x / 2+(x / 2)^3 / 3+(1 / 3)\left[(x / 2)+\cdot(x / 2)^3 / 3+\ldots\right)\right]^3-x}{\tan x-x}}

                                 {\Longrightarrow L=1+\lim _{x \rightarrow+\ldots 0} \frac{\left.2] x / 2+(x / 2)^3 / 3+(1 / 3)(x / 2)^3+O\left(x^4\right)\right]-x}{\tan x-x}}

                                 {\Longrightarrow L=1+\lim _{x \rightarrow+\ldots 0} \frac{\left(x+x^3 / 6+\ldots\right)-x}{x^3 / 3+O\left(x^5\right)}}

\text { Finally, we get }

                                 {L=\frac{3}{2}}

 

Posted by

Suraj Bhandari

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