Question

How do you use the limit definition to find the slope of the tangent line to the graph `y = 2^x` at (1,0)?

Answer 1

The point `(1,0)` does not lie on the graph of `y = 2^x`; here is the graph to prove it:

graph{2^x [-10, 10, -5, 5]}

Please observe that the y value approaches 0 but never actually reaches it.

Answer 2

slope `= ln 2 = 0.693147 ...`

Explanation:

Assuming the correct coordinate is `(0,1)`. The slope of the tangent at any particular point is given by the value of the derivative at that point. So we seek that value of `d/dx(2^x)` when `x=0`

The definition of the derivative of `y=f(x)` is

`dy/dx = f'(x) = lim_(h rarr 0) ( f(x+h)-f(x) ) / h`

So with `y=2^x` at the point `(0,1)` then;

`f'(0) = lim_(h rarr 0) ( f(h)-f(0) ) / h`
`\ \ \ \ \ \ \ \ = lim_(h rarr 0) ( 2^h-2^0 ) / h`
`\ \ \ \ \ \ \ \ = lim_(h rarr 0) ( 2^h - 1 ) / h`

This is as far as we can go using "conventional" analysis.

Investigation will show that the above limits indeed exists and converges to an irrational number approximately `0.693147 ...`. We discover that in fact this limit converges to `log_e 2` (or `ln2`) where `e=2.718281828459045...` is Euler's Number discovered by Leonhard Euler .

Thus we find:

`f'(0) = ln 2`
`\ \ \ \ \ \ \ \ = 0.693147 ...`