Question

how would I solve this limit ? : limx-->0 (2^x)-1/x

Answer 1

\[\lim_{x \rightarrow 0}\frac{ a ^{x}-1 }{x }=\log a\]

Answer 2

Let's think...
I see 2^x there
I know 2^0=1.
So...
Let's use that little thing we call the derivative:

\[f'(a)=\lim_{x \rightarrow a}\frac{2^x-2^a}{x-a}\]
Compare this to your expression.
what do you notice is f(x) here?

Answer 3

\[f(x)=b^x => f'(x)=b^x \cdot \ln(b) \]

Answer 4

how did you know to use that limit definition, and not f(x+h)-f(x)/h ?

Answer 5

I just looked for the definition that was prettiest to me use here.
You could use the other one here...
It just looked prettier to me to use that one
but since you want to involved that little change of x thing we can do that.
\[f'(a)=\lim_{h \rightarrow 0}\frac{2^{a+h}-2^{a}}{h}\]
Now remember h=x-a
So we should notice that we have a here is 0 and a+h=x.

Answer 6

wait why is h= x-a ?

Answer 7

ohhh nvm i see it

Answer 8

Because h is just a change in x
We took the always changing x variable along with a constant x variable such as a to note the change, h.

Answer 9

Like both of those formulas are the slope formula.
\[\lim_{h->0} \frac{f(a+h)-f(a)}{h}=\lim_{x->a}\frac{f(x)-f(a)}{x-a}\]
h=x-a (therefore h+a=x)
Now since h->0 and h+a=x and x=h+a is continuous at h equals 0 we can plug in 0 for h to find what x approaches
\[\lim_{h->0} (x)=\lim_{h->0} (h+a)=0+a=a\]

Answer 10

So x goes to a.

Answer 11

This might look familiar to you if you have evaluated limits using substitution.

Answer 12

That is all this is just rewriting a limit using a substitution.