Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis :

# If $$\lim_{x \rightarrow 0} \frac{ax-(e^{4x}-1)}{ax(e^{4x}-1)}$$ exists and is equal to b, then the value of a – 2b is

$$\begin{array}{l}\lim_{x \rightarrow 0} \frac{ax-(e^{4x}-1)}{ax(e^{4x}-1)}\end{array}$$

Applying L’Hospital Rule

$$\begin{array}{l}\lim_{x \rightarrow 0} \frac{a-4e^{4x}}{a(e^{4x}-1)+ax(4e^{4x})}\end{array}$$
So for limit to exist,a=4

Applying L’Hospital Rule

$$\begin{array}{l}\lim_{x \rightarrow 0} \frac{-16e^{4x}}{a(4e^{4x}) +a(4e^{4x})+ax(16e^{4x})}\end{array}$$

(-16)/(4a+4a)=(–16)/32=–1/2=b

a-2b = 4–2((–1)/2) = 4+1 = 5