Lim h-0 h/e^h-1 ?

1 Answer
Mar 16, 2018

1

Explanation:

We know that,
#color(red)(lim_(h to0)(a^h-1)/h=lna,anda=e=>lim_(h to0)(e^h-1)/h=lne=1)#
#L=lim_(h to0)h/(e^h-1)#
#=lim_(h to0)((h/h)/((e^h-1)/h))#, #(h to0=>h!=0)#
#=1/((lim_(h to0)(e^h-1)/h))=1/lne=1/1=1#