Determining a limit knowing partial derivative

Question

Let $f\in \mathcal C^1(\mathbb{R}^3,\mathbb{R})$ such that: $$f(0,1,1)=0, \quad \frac{\partial f}{\partial x}(0,1,1)=1, \quad \frac{\partial f}{\partial y}(0,1,1)=2, \quad\frac{\partial f}{\partial z}(0,1,1)=3.$$ Determine the limit of $$\lim_\limits{t\to0}\frac{f(t^2,\cosh t,e^t)}{f(t,\cos t,\cosh t)}$$

Obviously we have $f(t^2,\cosh t,e^t)(0) = f(0,1,1) = 0$ but I don't see the link between the derivative with the fraction.

Answer 1

$[1]:$ $\displaystyle \lim_{t \to 0} \dfrac{f(t^2,cosht,e^t)}{f(t,cost,cosht)}=\dfrac{0}{0} \to$ we need to apply L'Hôpital's rule.

Thus we have: $\displaystyle \lim_{t \to 0} \dfrac{f(t^2,cosht,e^t)}{f(t,cost,cosht)}=\lim_{t \to 0} \dfrac{d(f(t^2,cosht,e^t))}{d(f(t,cost,cosht))}$

$[2]:$ Also we know that: $df(x,y,z)=\dfrac{\partial f}{\partial x} dx + \dfrac{\partial f}{\partial y} dy + \dfrac{\partial f}{\partial z} dz$

Hence:

$d(f(t^2,cosht,e^t))=\dfrac{\partial f}{\partial x} dx + \dfrac{\partial f}{\partial y} dy + \dfrac{\partial f}{\partial z} dz=\dfrac{\partial f}{\partial x} 2t dt + \dfrac{\partial f}{\partial y} \sinh{t} dt + \dfrac{\partial f}{\partial z} e^t dt$

$d(f(t,cost,cosht))=\dfrac{\partial f}{\partial x} dx + \dfrac{\partial f}{\partial y} dy + \dfrac{\partial f}{\partial z} dz=\dfrac{\partial f}{\partial x} dt + \dfrac{\partial f}{\partial y} (-\sin t) dt + \dfrac{\partial f}{\partial z} \sinh t dt$

Thus we have: $\displaystyle \lim_{t \to 0} \dfrac{f(t^2,cosht,e^t)}{f(t,cost,cosht)}=\lim_{t \to 0} \dfrac{d(f(t^2,cosht,e^t))}{d(f(t,cost,cosht))}=\dfrac{\dfrac{\partial f}{\partial x}(0,1,1) 2(0) dt + \dfrac{\partial f}{\partial y}(0,1,1) \sinh{0} dt + \dfrac{\partial f}{\partial z}(0,1,1) e^0 dt}{\dfrac{\partial f}{\partial x}(0,1,1) dt + \dfrac{\partial f}{\partial y}(0,1,1) (-\sin 0) dt + \dfrac{\partial f}{\partial z}(0,1,1) \sinh 0 dt}=\dfrac{\dfrac{\partial f}{\partial z}(0,1,1) e^0}{\dfrac{\partial f}{\partial x}(0,1,1) }=\frac{3}{1}=3$

Answer 2

\begin{align*} &\lim_{t\rightarrow 0}\dfrac{f(t^{2},\cosh t,e^{t})}{f(t,\cos t,\cosh t)}\\ &=\lim_{t\rightarrow 0}\dfrac{f(0,1,1)+\left(\dfrac{\partial f}{\partial x}(0,1,1),\dfrac{\partial f}{\partial y}(0,1,1),\dfrac{\partial f}{\partial z}(0,1,1)\right)\cdot(t^{2},\cosh t-1,e^{t}-1)}{f(0,1,1)+\left(\dfrac{\partial f}{\partial x}(0,1,1),\dfrac{\partial f}{\partial y}(0,1,1),\dfrac{\partial f}{\partial z}(0,1,1)\right)\cdot(t,\cos t -1,\cosh t-1)}\\ &=\lim_{t\rightarrow 0}\dfrac{t^{2}+2(\cosh t-1)+3(e^{t}-1)}{t+2(\cos t-1)+3(\cosh t-1)}\\ &=\lim_{t\rightarrow 0}\dfrac{2t+2\sinh t+3e^{t}}{1-2\sin t+3\sinh t}\\ &=3. \end{align*}