limit of (e^x -1)/2x as x tends to 0?

2 Answers
Jan 16, 2018

#lim_(x->0)(e^x-1)/(2x)=1/2#

Explanation:

I'll assume you meant #lim_(x->0)(e^x-1)/(2x)#

Investigate the limit first by using direct substitution:

#lim_(x->0)(e^x-1)/(2x)=(e^color(red)0-1)/(2*color(red)(0))=(1-1)/0=0/0#

We get the indeterminate form #0/0# which should indicate that we should use L'hospital's Rule to rewrite the limit into something we can evaluate

L'Hospital states:

If #lim_(x->a)f(x)/(g(x))=0/0#

Then, #lim_(x->a)f(x)/g(x)=lim_(x->a)(d/dx[f(x)])/(d/dx[g(x)])#

In this example:

#f(x)=e^x-1=>d/dx[e^x-1]=e^x#

#g(x)=2x=>d/dx[2x]=2#

#lim_(x->0)(e^x-1)/(2x)=lim_(x->0)(d/dx[e^x-1])/(d/dx[2x])=lim_(x->0)e^x/2#

Now we can attempt to evaluate the limit:

#lim_(x->0)e^x/2=e^color(red)0/2=1/2#

Jan 16, 2018

We can say it is around #1/2#

Explanation:

I am just trying to visualize what is going on here.

Let's graph the function #f(x)=(3^x-1)/(2x)#
graph{(3^x-1)/(2x) [-10, 10, -5, 5]}

We can see that the function is continuous except at #x=0#.
We also can see that as #x# approaches 0, it is getting really close to #1/2# from both the positive and the negative directions.