How do you find #lim (e^t-1)/t# as #t->0# using l'Hospital's Rule?

2 Answers
Hammer
Jun 1, 2018

#lim_(t->0)(e^t-1)/t=1#

Explanation:

We have

#L=lim_(t->0)(e^t-1)/t#

To apply L'Hôpital's rule, we must have a #0"/"0# or #oo"/"oo# situation. If we plug in #t=0# we find that:

#L=(e^0-1)/0=0/0#

Thus we can use L'Hôpital's rule, which states that:

#L=lim_(t->0)(e^t-1)/t=("d"/("d"t) (e^t-1))/("d"/("d"t)t#

We know that #e^x# is one of the functions with the property that #f'(x)=f(x)#, and as #-1# is just a constant, it will vanish when we take the derivative.

#:. L=lim_(t->0)color(red)(e^t)/color(red)1=lim_(t->0)color(red)(e^t)#

#color(red)(L=e^0=1)#

Jim S
Jun 1, 2018

#lim_(trarr0)(e^t-1)/t=1#

Explanation:

#lim_(trarr0)(e^t-1)/t=_(DLH)^((0/0))lim_(trarr0)e^t=e^0=1#

Related questions
See all questions in Indeterminate Forms and de L'hospital's Rule
Impact of this question
20093 views around the world
You can reuse this answer
Creative Commons License