Answer
$\lim\limits_{\Delta x \to 0}\dfrac{(x+\Delta x)^{2}-x^{2}}{\Delta x}=2x$
Work Step by Step
$\lim\limits_{\Delta x \to 0}\dfrac{(x+\Delta x)^{2}-x^{2}}{\Delta x}$
Applying direct substitution immediately will result in an indeterminate form. So first, evaluate $(x+\Delta x)^{2}$:
$\lim\limits_{\Delta x \to 0}\dfrac{(x+\Delta x)^{2}-x^{2}}{\Delta x}=\lim\limits_{\Delta x \to 0}\dfrac{x^{2}+2x\Delta x+\Delta x^{2}-x^{2}}{\Delta x}=...$
$...=\lim\limits_{\Delta x \to 0}\dfrac{2x\Delta x+\Delta x^{2}}{\Delta x}=...$
Take out common factor $\Delta x$ in the numerator and simplify:
$...=\lim\limits_{\Delta x \to 0}\dfrac{\Delta x(2x+\Delta x)}{\Delta x}=\lim\limits_{\Delta x \to 0}2x+\Delta x=...$
Now, apply direct substitution to evaluate the limit:
$\lim\limits_{\Delta x \to 0}2x+\Delta x=2x+0=2x$
