Calculating $\lim_{x\to\infty}(a^x+b^x-c^x) \ ; \lim_{x\to0}(a^x+b^x-c^x)$

Question

Calculate: $\displaystyle\lim_{x\to\infty}(a^x+b^x-c^x) \ ; \lim_{x\to0}(a^x+b^x-c^x) \ :\forall a>b>c>0$.

First of all, is it actually necessary to check all 5 possibilities for $c=1, c>1,b=1,a=1,a>1$ ?

I also run into trouble for $a=1$, I chose some arbitrary numbers to make it easier to see:

$\displaystyle\lim_{x\to\infty}(1+(\frac 1 2)^x-(\frac 1 3)^x)= 1+0-0=1$ but if I make a small change by dividing by $(\frac 1 3)^x$: $\displaystyle\lim_{x\to\infty}(3^x+(\frac 3 2)^x-1)=\infty\neq 1$, so why it's apparently wrong to divide by smallest element here ?

Answer 1

Let $B=b/a<1$ and $C=c/a<1$ $$ a^x+b^x-c^x=a^x(1+B^x-C^x) $$ What is $\lim_{x\to\infty}r^x$ when $0<r<1$?

Now divide in the cases $0<a<1$, $a=1$ and $a>1$.

For the limit at $0$ is there any problem in showing it is $1$?