## University Calculus: Early Transcendentals (3rd Edition)

The equation of the tangent line is $y=x+1$.
The slope $m$ of the tangent line of the curve $f(x)$ at $A(a,b)$ is also the slope of the curve at that point, which is calculated by $$m=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$$ $$y=f(x)=2\sqrt x\hspace{1cm}A(1,2)$$ 1) Find the slope $m$ of the tangent: $$m=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$$ Here $a=1$ and $f(a)=b=2$. $$m=\lim_{h\to0}\frac{2\sqrt{1+h}-2}{h}=2\lim_{h\to0}\frac{\sqrt{1+h}-1}{h}$$ Multiply both numerator and denominator with $\sqrt{1+h}+1$: $$m=2\lim_{h\to0}\frac{(\sqrt{1+h}-1)(\sqrt{1+h}+1)}{h(\sqrt{1+h}+1)}$$ $$m=2\lim_{h\to0}\frac{1+h-1}{h(\sqrt{1+h}+1)}=2\lim_{h\to0}\frac{h}{h(\sqrt{1+h}+1)}=2\lim_{h\to0}\frac{1}{\sqrt{1+h}+1}$$ $$m=2\times\frac{1}{\sqrt{1+0}+1}=\frac{2}{\sqrt1+1}=1$$ 2) Find the equation of the tangent line at $A(1,2)$: The tangent line would have this form: $$y=x+m$$ Substitute $A(1,2)$ here to find $m$: $$1+m=2$$ $$m=1$$ So the equation of the tangent line is $y=x+1$.