To solve the limit
‌, let's begin by rewriting each term in the numerator using the exponential function. Recall that for any real number
y, the expression
ay can be expressed as
ey‌ln‌a.
The expression thus becomes:
‌| ex‌ln‌a−ex‌ln‌b |
| x |
.We can further rewrite this using the first few terms of the Taylor expansion for
eu around
u=0, which is
eu≈1+u when
u is small. Substituting
u=x‌ln‌a for the first term and
u=x‌ln‌b for the second term, we get:
‌ex‌ln‌a≈1+x‌ln‌a‌ex‌ln‌b≈1+x‌ln‌bSubstituting these approximations into the limit, we have:
‌| (1+x‌ln‌a)−(1+x‌ln‌b) |
| x |
Expand and simplify the expression in the numerator:
(1+x‌ln‌a)−(1+x‌ln‌b)=x‌ln‌a−x‌ln‌b=x(ln‌a−ln‌b)‌. ‌Now, the limit simplifies to:
‌‌. ‌Since
x in the denominator and numerator cancel out, we are left with:
ln‌a−ln‌b.Using the properties of logarithms, specifically
ln‌a−ln‌b=ln‌, the expression simplifies to:
ln‌‌. ‌Therefore, the answer to the problem
‌ is indeed:
ln‌,which corresponds to Option C.