Question

If lim$_{x\to 0} \frac{ax - (e^{4x}-1)}{ax(e^{4x}-1)}$ exists and is equal to b, then the value of a$-$2b is ________ .

Hint:

When evaluating limits of the form 0/0, using Taylor series expansions is a very robust method. Expand the functions around the limit point and equate coefficients of the lowest powers of x to satisfy the condition of a finite, non-zero limit.

Answer 1

Correct Answer: 5

\[ \lim_{x\to0}\frac{ax-(e^{4x}-1)}{ax(e^{4x}-1)} \] Use expansion: \[ e^{4x}-1=4x+8x^2+O(x^3) \] Numerator: \[ (a-4)x-8x^2 \] Denominator: \[ 4ax^2 \] For limit to exist: \[ a-4=0 \Rightarrow a=4 \] \[ b=\lim_{x\to0}\frac{-8x^2}{16x^2}=-\frac12 \] \[ a-2b=4+1=5 \] \[ \boxed{5} \]