Question

If `lim_(x rightarrow 0) (ax - (e^(4x) - 1))/(ax(e^(4x) - 1))` exists and is equal to b, then the value of a – 2b is  ______.

Options

• 4

• 5

• 6

• 7

Answer 1

If `lim_(x rightarrow 0) (ax - (e^(4x) - 1))/(ax(e^(4x) - 1))` exists and is equal to b, then the value of a – 2b is 5.

Explanation:

`lim_(x rightarrow 0) (ax - e^(4x) + 1)/(ax(e^(4x) - 1))`, at `x rightarrow 0(0/0)`

Applying L'Hospital rule

= `lim_(x rightarrow 0) (a - 4e^(4x))/((e^(4x) - 1) + 4axe^(4x))`

At x = 0, `(a - 4)/0`

∵ Limits exists only when a – 4 = 0

So `0/0` case

Applying L'Hospital rule

= `lim_(x rightarrow 0) (-16e^(4x))/(4ae^(4x) + 4ae^(4x) + 16axe^(4x))`

⇒ `(-16)/(8a)` = b

⇒ `(-16)/(8.4)` = b

⇒ b = `-1/2`

So, a – 2b = 4 + 1 = 5