--- ## Question 3: Define two forms of the limit definition of a derivative. --- #### Form 1: ## $$\lim_{h\to 0}f(x)=\frac{f(x+h)-f(x)}{x+h}$$ --- ### Using the equation: $$x^2 +1$$ #### $$\lim_{h\to 0}f(x)=\frac{((x+h)^2 +1)-(x^2 +1)}{h}$$ --- #### This next step we simplify the equation by factoring everything out. ### $$\lim_{h\to 0}f(x)=\frac{x^2 +2xh+h^2 -x^2 +1}{h}$$ #### Next we canceled out like terms in the equation. ### $$\lim_{h\to 0}f(x)=\frac{2xh+h^2 }{h}$$ #### Then we can factor out an h since it is a like term. #### $$\lim_{h\to 0}f(x)=\frac{h(2x+h)}{h}$$ --- #### H is then canceled since it has an h on the numerator and denominator, leaving us with: ### $$\lim_{h\to 0}f(x)=2x+h$$ ### Then h gets set to 0 leaving us with: ### $$\lim_{h\to 0}f(x)=2x$$ --- ## Form 2: ### Definition:$$\lim_{x\to a}=\frac{f(x)-f(a)}{x-a}$$ ### Using the equation $$x^2 +1$$ ## $$\lim_{x\to a}=\frac{(x^2 +1)-(a^2 +1)}{x-a}$$ --- #### We have to apply L'Hospital's rule since the expression is undefined: ### $$\lim_{x\to a}=\frac{\frac{d}{dx}(x^2 +1)-(a^2 +1)}{\frac{d}{dx}x-a}$$ ### Taking the derivative of the numerator and denominator: #### $$\lim_{x\to a}=\frac{2x+0}{1+0}=2x$$ ---