If the function ƒ defined on ( - 1/3,1/3 ) by
f(x) = { 1/xlog ₑ( (1 + 3x)/(1 - 2x) ),, when x ≠ 0; k,, when x = 0
is continuous, then k is equal to_______.

Solution

Maths sir · 4 steps

Split the log of a quotient into two logs, and each piece becomes the standard limit ln(1+t)/t = 1.

1. What continuity needs. For f to be continuous at x = 0, the value k must equal lim(x → 0) f(x). So we only have to find that limit.
2. Split the log. Write ln((1+3x)/(1-2x)) = ln(1+3x) - ln(1-2x). Then divide each piece by x and fix the denominators so they match the inside of each log.

   lim(x → 0) f(x) = 3 · lim(x → 0) ln(1+3x)/3x + 2 · lim(x → 0) ln(1-2x)/-2x = 3(1) + 2(1)

3. Use the standard limit. Each fraction has the form ln(1+t)/t with t → 0, so each tends to 1. The limit is 3 + 2 = 5.
4. So. Continuity forces k = lim(x → 0) f(x), so k = 5.

The common slip: 1. Mishandling the minus sign on ln(1-2x) makes it 3 - 2 = 1 instead of 3 + 2 = 5.

The idea to keep. Whenever you see 1/x ln(something near 1), split the log and turn each part into ln(1+t)/t → 1. Watch the signs: ln(1-2x) contributes -2, and subtracting it gives +2.

Answer from the NTA final answer key. Solution by PrepWiser.