Use the limit definition to find the slope of the tangent to f(x)=x² at (1,1)

Core Idea: Here the point is not to memorize 2x, but to show how the difference quotient collapses to the derivative rule after algebra and the h → 0 step.

What Is This Asking?

The note asks for the tangent slope to f(x)=x² at (1,1) and walks through the limit definition of the derivative with explicit algebra.

Step-by-Step Solution

Setup. The slope of the tangent line at a point is the derivative at that point. From the limit definition:

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

Take $f(x)=x^2$. Then

$\frac{(x+h)^2-x^2}{h}=\frac{x^2+2xh+h^2-x^2}{h}=\frac{2xh+h^2}{h}=2x+h\quad (h\neq 0).$

Let $h\to 0$: $f'(x)=2x$.

At $(1,1)$. The slope of the tangent line is $f'(1)=2$.

Interpretation. For this parabola, the tangent line at $x=1$ rises 2 units vertically per 1 unit horizontally (until you leave the local linear approximation).

Pitfall Alert

Canceling $h$ without explicitly assuming $h\neq 0$ is a common write-up gap. Also, don’t plug $h=0$ into $(2xh+h^2)/h$ before simplifying — simplify the quotient first, then take the limit.

Try a Variant

If the question were $f(x)=x^3$, the same algebra yields $f'(x)=3x^2$, so at $x=1$ the slope would be $3$. If the point moved to $x=a$, the slope would be $2a$ for $f(x)=x^2$.

Further Reading

difference quotient, derivative, tangent line, limit, power rule