Question

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

Answer 1

`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`

Answer 2

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.