Recall that the derivative of a function f(x) at a point x = c is given by
[tex]\displaystyle f'(c) = \lim_{x\to c} \frac{f(x) - f(c)}{x - c}[/tex]
By substituting h = x - c, we have the equivalent expression
[tex]\displaystyle f'(c) = \lim_{h\to0} \frac{f(c+h) - f(c)}h[/tex]
since if x approaches c, then h = x - c approaches c - c = 0.
The two given limits strongly resemble what we have here, so it's just a matter of identifying the f(x) and c.
For the first limit,
[tex]\displaystyle \lim_{h\to0} \frac{\sin\left(\frac\pi3 + h\right) - \frac{\sqrt3}2}h[/tex]
recall that sin(π/3) = √3/2. Then c = π/3 and f(x) = sin(x), and the limit is equal to the derivative of sin(x) at x = π/3. We have
[tex](\sin(x))' = \cos(x)[/tex]
and cos(π/3) = 1/2.
For the second limit,
[tex]\displaystyle \lim_{a\to0} \frac{e^{2a} - 1}a[/tex]
we observe that e²ˣ = 1 if x = 0. So this limit is the derivative of e²ˣ at x = 0. We have
[tex]\left(e^{2x}\right)' = e^{2x} (2x)' = 2e^{2x}[/tex]
and 2e⁰ = 2.
