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What should be the value of f(0) so that the function \mathrm{f(x)=\frac{e^{\tan x}-e^x+\ln (\sec x+\tan x)-x}{\tan x-x} } is continuous at x=0 ?

Option: 1

\frac{1}{8}


Option: 2

1.6


Option: 3

1


Option: 4

1.5


Answers (1)

best_answer

For f(x) to be continuous at x=0, we must have
\mathrm{\begin{aligned} & \lim _{x \rightarrow 0} \frac{e^{\tan x}-e^x+\ln (\sec x+\tan x)-x}{\tan x-x}=f(0) \\ & \Rightarrow f(0)=\lim _{x \rightarrow 0} e^x \frac{\left(e^{\tan x-x}-1\right)}{\tan x-x}+\lim _{x \rightarrow 0} \frac{\ln (\sec x+\tan x)-x}{\tan x-x} \end{aligned} }
\mathrm{\begin{aligned} & \Rightarrow f(0)=\lim _{x \rightarrow 0} e^x \frac{\left(e^{\tan x-x}-1\right)}{\tan x-x}+\lim _{x \rightarrow 0} \frac{\ln (\sec x+\tan x)-x}{x^3\left(\frac{\tan x-x}{x^3}\right)} \\ & \Rightarrow f(0)=1 \times e^0+3 \lim _{x \rightarrow 0} \frac{\ln (\sec x+\tan x)-x}{x^3} \\ & \Rightarrow f(0)=1+3 \lim _{x \rightarrow 0} \frac{\sec x-1}{3 x^2} \quad\left[\lim _{x \rightarrow 0} \frac{\tan x-x}{x^3}=\frac{1}{3}\right] \\ & \Rightarrow f(0)=1+3 \lim _{x \rightarrow 0} \frac{1-\cos x}{3 \cos x \cdot x^2} \\ & \Rightarrow f(0)=1+3 \times \frac{1}{3} \times \frac{1}{2}=\frac{3}{2} \end{aligned} }
[By L' Hospital rule]

Posted by

Pankaj Sanodiya

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