Hi!

Is my solution correct?

Let $\displaystyle f(x) = x^2 -1$. Find the equation of the line tangent to the graph of $\displaystyle f$ at the point where $\displaystyle x = 2$.

$\displaystyle \lim_{h \to 0} = \frac{f(x+h) - f(x)}{h}$

$\displaystyle \lim_{h \to 0} = \frac{f(2+h)^2 - f(2)}{h}$

$\displaystyle \lim_{h \to 0} = \frac{\left[(2+h)^2 - 1\right] - (2^2-1)}{h}$

$\displaystyle \lim_{h \to 0} = \frac{h^2 + 4h + 4 - 4 + 1}{h}$

$\displaystyle \lim_{h \to 0} = \frac{h^2 + 4h + 1}{h}$

$\displaystyle \lim_{h \to 0} = \frac{h(h+4) - 3}{h}$

$\displaystyle \lim_{h \to 0} (h+4) = -3$

Limit at $\displaystyle x=2$ doesn't exist.