Using the Euler's method (with $h={10}^{-n}$) to find $y(1)$

${y}^{\prime}=\frac{\mathrm{sin}(x)}{x}$

Since

${y}^{\prime}(x)=\frac{\mathrm{sin}(x)}{x}$

then

$f(x,y)=\frac{\mathrm{sin}(x)}{x}$

I know

${y}_{n+1}={y}_{n}+h\cdot f({x}_{n},{y}_{n})$

and given $y(0)=0$, so

${x}_{0}={y}_{0}=0$

Therefore,

${x}_{1}={x}_{0}+h=0+{10}^{-0}=1\Rightarrow {y}_{1}=y({x}_{1})=y(1)$

Then,

${y}_{1}=0+{10}^{-0}\cdot f({x}_{0},{y}_{0})$

${y}_{1}=f(0,0)$

But

$f(0,0)=\frac{\mathrm{sin}(0)}{0}=\frac{0}{0}$

How am I supposed to do it?

${y}^{\prime}=\frac{\mathrm{sin}(x)}{x}$

Since

${y}^{\prime}(x)=\frac{\mathrm{sin}(x)}{x}$

then

$f(x,y)=\frac{\mathrm{sin}(x)}{x}$

I know

${y}_{n+1}={y}_{n}+h\cdot f({x}_{n},{y}_{n})$

and given $y(0)=0$, so

${x}_{0}={y}_{0}=0$

Therefore,

${x}_{1}={x}_{0}+h=0+{10}^{-0}=1\Rightarrow {y}_{1}=y({x}_{1})=y(1)$

Then,

${y}_{1}=0+{10}^{-0}\cdot f({x}_{0},{y}_{0})$

${y}_{1}=f(0,0)$

But

$f(0,0)=\frac{\mathrm{sin}(0)}{0}=\frac{0}{0}$

How am I supposed to do it?