Continuity

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$f\left(x\right)=\{\begin{array}{l}\frac{lo{g}_e\left(1-x+x^2\right)+lo{g}_e\left(1+x+x^2\right)}{secx-cosx},x\in \left(\frac{-\pi }{2},\frac{\pi }{2}\right)-\left\{0\right\}\\ k,x=0\end{array}$ for continuity at x = 0

      $\underset{x\to 0}{lim}f\left(x\right)=k\therefore k=\underset{x\to 0}{lim}\frac{lo{g}_e\left(1+x^2+x^4\right)}{secx-cosx}\left(\frac{0}{0}form\right)=\underset{x\to 0}{lim}\frac{cosxlo{g}_e\left(1+x^2+x^4\right)}{si{n}^{2}x}=1$  

RHL&LHL lim (x→0) (sin (2x²/a) + cos (3x/b)^ (ab/x²)
= e^ (lim (x→0) (sin (2x²/a) + cos (3x/b) - 1) (ab/x²) = e^ (4b²-9a)/2b)
f (0) = e³
For continuity at x = 0
Limit = f (0)
(4b² - 9a)/2b = 3 ⇒ 4b² – 6b – 9a = 0∀b ∈ R
⇒ D ≥ 0 ⇒ a ≥ -1/4
a? = -1/4
⇒ |1/a? | = 4.

LHL = lim (x→2? ) λ|x²-5x+6| / µ (5x-x²-6) = lim (x→2? ) -λ/µ
RHL = lim (x→2? ) e^ (tan (x-2)/ (x- [x]) = e¹
f (2) = µ
For continuity, -λ/µ = e = µ.
⇒ µ=e, -λ=µ²=e², λ=-e²
∴ λ + µ = -e² + e = e (1-e)