Limit Question: I have completed this problem by myself, but I want to confirm that I did the problem correctly

Question

Sorry for posting this question in image format, I was having a hard time posting the problem here.

Answer 1

You should first start with the continuity condition.

In order for function to be continuous we must have $$\lim_{x\to 1^{-}}g(x)=g(1)=\lim_{x\to 1^{+}}g(x)$$ so that $$a+3=b=c+2$$ Next it's time to use differentiability condition. Here you have made a subtle mistake. You make the assumption of the derivative $g'$ being continuous at $1$. The right approach is to use definition of derivative.

We have $$g'_{+} (1)=\lim_{x\to 1^{+}}\frac{g(x)-g(1)}{x-1}=\lim_{x\to 1^{+}}\frac{cx^3+2-b}{x-1}$$ Putting $b=c+2$ in above limit we can see that $g'_{+} (1)=3c$. Similarly we have $$g'_{-} (1)=\lim_{x\to 1^{-}}\frac{3x^2+a-b}{x-1}$$ Putting $b=a+3$ we can see that $g'_{-}(1)=6$. In order for derivative to exist we must have these two limits equal so that $3c=6$. Therefore $c=2,b=4,a=1$.