Find all parameters $a,b,c,d \in \mathbb R$ for which the function $f: \mathbb R \rightarrow \mathbb R$ a) has a limit at point 0 b) is continuous My function $f: \mathbb R  \rightarrow \mathbb R$ it is given a pattern:
$$f(x)=\begin{cases}  \frac{2^{7^{x}-1}-a}{\ln(1-x)},& x<0\\ b,& x=0 \\ \frac{\sin\left(c \sqrt{x^{2}+d^{2}x}\right) }{x},& x>0  \end{cases}$$ for $$a,b,c,d \in \mathbb R$$
I have a problem with this task because function patterns for $x</>0$ have they reach zero in the denominator for $x=0.$
That is why I think that I should cut a denominator with a numerator but I cannot do it because I have logarithm and sine so I need other way to this task.
I thought about doing some substitution: for example $y=\frac{1}{x}$, then for $x \rightarrow 0^\mp$ I have $y \rightarrow \pm \infty$ and but I still can not solve this.
Can I count on any tips?

The first thing to note is that as $x\nearrow 0,$ the denominator $\ln(1-x)$ vanishes, so the only way to avoid $\bigl|f(x)\bigr|\to+\infty$ in that case is to make sure that the numerator vanishes as well. Since $$2^{7^x-1}-a\to2^{7^0-1}-a=2^{1-1}-a=2^0-a=1-a,$$ this means we must put $a=1$ to accomplish this.
If you're familiar with the result that $$\lim_{t\to 0}\frac{\sin(t)}t=1,\tag{1}$$ then we can tackle the right-hand limit as follows. Since $\lim_{x\searrow0}c\sqrt{x^2+d^2x}=0,$ then by $(1)$ we have $$\lim_{x\searrow0}\frac{\sin\left(c\sqrt{x^2+d^2x}\right)}{c\sqrt{x^2+d^2x}}=1,$$ so, since $$\frac{\sin\left(c\sqrt{x^2+d^2x}\right)}{x}=\frac{\sin\left(c\sqrt{x^2+d^2x}\right)}{c\sqrt{x^2+d^2x}}\cdot\frac{c\sqrt{x^2+d^2x}}{x}=\frac{\sin\left(c\sqrt{x^2+d^2x}\right)}{c\sqrt{x^2+d^2x}}\cdot c\sqrt{1+\frac{d^2}x}$$ whenever $x>0,$ then $\lim_{x\searrow0}f(x)=+\infty$ unless $d=0,$ in which case the limit is $c.$ Thus, for the right-hand limit to exist, we require $d=0.$ Moreover, for the two-sided limit to exist, we need $a=1,$ $d=0,$ and $c=\lim_{x\nearrow0}f(x),$ assuming the left-hand limit exists.
Unfortunately, I'm not familiar with any elementary ways to proceed further. Using L'Hôpital's rule allows us to find that $$\begin{eqnarray}\lim_{x\nearrow0}\frac{2^{7^x-1}-1}{\ln(1-x)} &=& \lim_{x\nearrow0}\cfrac{\ln(2)\ln(7)7^{x}2^{7x-1}}{-\frac1{1-x}}\\ &=& \lim_{x\nearrow0}(x-1)\ln(2)\ln(7)7^{x}2^{7x-1}\\ &=& (0-1)\ln(2)\ln(7)7^{0}2^{0-1}\\ &=& -\frac{\ln(2)\ln(7)}2,\end{eqnarray}$$ so for the two-sided limit to exist, we require $a=1,$ $c=\frac{\ln(2)\ln(7)}2,$ and $d=0.$ For continuity, we additionally require $b=\frac{\ln(2)\ln(7)}2.$
