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

Question

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

anonymous · Answer

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

myininaya · Answer

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?

myininaya · Answer

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

anonymous · Answer

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

myininaya · Answer

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.

anonymous · Answer

wait why is h= x-a ?

anonymous · Answer

ohhh nvm i see it

myininaya · Answer

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.

myininaya · Answer

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
$$\lim_{h->0} (x)=\lim_{h->0} (h+a)=0+a=a$$

myininaya · Answer

So x goes to a.

myininaya · Answer

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

myininaya · Answer

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