An interesting question.
Find the equation of the tangent for the curve $\displaystyle y = x^2 $ at the point $\displaystyle (1,1)$ without using calculus.
you mean without using differentiation, right? well, if so, suppose the line is $\displaystyle y=mx + b.$ since (1,1) belongs to the line, we have $\displaystyle b=1-m.$ now the tangent line should intersect the curve
at one point only, i.e. the equation $\displaystyle x^2 - mx + m-1=0$ must have only one solution. so the discriminant has to be 0, which gives us $\displaystyle m=2.$ thus the line is $\displaystyle y=2x-1.$
since x = 1 must be the root of the equation $\displaystyle x^n-mx+m-1=0$ of multiplicity at least 2, we'll have: $\displaystyle x^n - mx + m - 1=(x-1)^2p(x),$ for some polynomial $\displaystyle p(x).$ thus:
$\displaystyle (x-1)(x^{n-1}+ \cdots + x + 1 - m)=(x-1)^2p(x),$ and hence: $\displaystyle x^{n-1} + \cdots + x + 1 - m = (x-1)p(x).$ not put $\displaystyle x=1$ to get $\displaystyle m=n,$ and therefore $\displaystyle b=1-n.$ so the equation
of the tangent line is $\displaystyle y=nx + 1-n.$
1. $\displaystyle \sin(x)$ at $\displaystyle (0,0)$. We know the tangent line is $\displaystyle \hat{y}-0=\cos(0)(x-0)\implies \hat{y}=x$
Ok so taking a cue from NonCommAlg and assuming that you mean no derivatives this is simple. It can be easily show using non caluculus terms that the function $\displaystyle \frac{\sin(x)}{x}$ approaches one as x approaches zero. So now consider this let $\displaystyle f$ be a differentiable function and $\displaystyle \hat{y}$ its tangent line at a point $\displaystyle x=c$. Consider that a tangent line approximates a curve in a neighborhood of $\displaystyle c$, and that $\displaystyle \hat{y}(c)=f(c)$. So it is obvious that $\displaystyle \frac{f(x)}{\hat{y}}$ must get closer and closer to one as x gets closer and closer to c. So we must have that $\displaystyle \frac{\sin(x)}{mx}$ must get closer and closer to one as x gets closer and closer to zero.But because as we noted $\displaystyle \frac{\sin(x)}{x}\to 1$ we must have that $\displaystyle \frac{\sin(x)}{mx}\to 1 \implies m=1$ so the tangent is $\displaystyle y=x$.
I implictly used calculus I suppose.
I think for the second one I can do better
For the function $\displaystyle f(x)=e^x$we know that the tangent line, $\displaystyle \hat{y}$, at $\displaystyle (0,1)$ must be of the form $\displaystyle \hat{y}-1=mx\Rightarrow \hat{y}=mx+1$. Now consider the tangent line $\displaystyle \hat{y}_2$ of the function $\displaystyle f_1(x)=e^{-x}$ at $\displaystyle (0,1)$. It is clear geometrically that these functions have perpendicular slopes, so the tangent line must be of the form $\displaystyle \hat{y}_2=\frac{-x}{m}+1$. But now consider that $\displaystyle f_1(-x)=f(x)$ in this case so $\displaystyle \hat{y}(x)=mx+1=\frac{x}{m}+1=\hat{y}_2(-x)\implies m=1$
Im not too sure about that one...can someone verify?