Find the value of limx→ 0ax−1x
We have,limx→ 0ax−1x
It is an examle of 00 form
While evaluating the limit,we can use expansion of ax.
We know,
ax=1+x(logea)+x22!(logea)2+......
substituting the expansion of ax in the numerator of the limit
=limx→ 0[1+x(logea)+x22!(logea)2+.......]−1x
=limx→ 0logea+x2!(logea)2+......
=logea+0
=logea