Mathematics - Continuity - Differentiability Question with Solution | TestHub

If $f\left(x\right)$ is continuous at $x=2$, then $\underset{x\to 2^{+}}{\lim }f\left(x\right)=\underset{x\to 2^{-}}{\lim }f\left(x\right)=\underset{x\to 2}{\lim }f\left(x\right)$

Here,

$\underset{x\to 2^{+}}{\lim }f\left(x\right)=\underset{x\to 2^{+}}{\lim }{e}^{\frac{\tan (x-2)}{x-2}}=e^1=e$

$\underset{x\to 2^{-}}{\lim }f\left(x\right)=\underset{x\to 2^{-}}{\lim }\frac{\lambda \left|x^2-5x+6\right|}{\mu \left(5x-x^2-6\right)}=\underset{x\to 2^{-}}{\lim }\frac{-\lambda (x-2)(x-3)}{\mu (x-2)(x-3)}=-\frac{\lambda }{\mu }$

$\left(\because \left|x^2-5x+6\right|=\left\{\begin{array}{l}\left(x-2\right)\left(x-3\right) \text{if x < 2 or x > 3}\\ -\left(x-2\right)\left(x-3\right) \text{if 2 < x < 3 }\end{array}\right.\right)$

For continuity $\mu =e=-\frac{\lambda }{\mu }\Rightarrow \mu =e, \lambda =-e^2$

$\lambda +\mu =e(-e+1)$