If the function f(x)=e^x(e^ x-x-1)+ _e( x+ x)-x x-x is continuous at x=0, then the value o…

Question

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

Accepted Answer

Given that $f(x)$ is continuous at $x=0$ means $\underset{x	o 0}{lim}f(x)=f(0)$

$f(0)=\underset{x	o 0}{lim}\frac{{e}^{x}({e}^{	an x-x}-1)+{\log }_{e}(\sec x+	an x)-x}{	an x-x}$

$f(0)=\underset{x	o 0}{lim}\frac{{e}^{x}({e}^{	an x-x}-1)}{	an x-x}+\underset{x	o 0}{lim}\frac{{\log }_{e}(\sec x+	an x)-x}{	an x-x}$

Using standard limit $\frac{{e}^{f(x)}-1}{f(x)}=1$ if $f(x)	o 0$

Let $f(x)=	an x-x$, then $\underset{x	o 0}{lim}\frac{{e}^{	an x-x}-1}{	an x-x}=1$

Second limit is $\frac{0}{0}$ form use L'Hopital Rule

$f(0)=1+\underset{x	o 0}{lim}\frac{\frac{d}{dx}(\log (\sec x+	an x)-x)}{\frac{d}{dx}(	an x-x)}$

$\begin{gathered} =1+\underset{x	o 0}{lim}\frac{\frac{1}{\sec x+	an x}(\sec x	an x+{\sec }^{2}x)-1}{{\sec }^{2}x-1} \ =1+\underset{x	o 0}{lim}\frac{\sec x	an x+{\sec }^{2}x-\sec x-	an x}{(\sec x+	an x)({\sec }^{2}x-1)} \ =1+\underset{x	o 0}{lim}\frac{	an x(\sec x-1)+\sec x(\sec x-1)}{(\sec x+	an x)({\sec }^{2}x-1)} \ =1+\underset{x	o 0}{lim}\frac{(\sec x-1)(\sec x+	an x)}{(\sec x+	an x)({\sec }^{2}x-1)} \ =1+\underset{x	o 0}{lim}\frac{1}{\sec x+1} \end{gathered}$

$\Rightarrow$ $f(0)=1+\frac{1}{2}=\frac{3}{2}$

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