Showing that $c^x - (c+c^{-x})^x \sim -x/c$

Question

Put $f_c(x) = c^x - (c+c^{-x})^x$ with $c>1$ a real number. I would like to show that $f_c(x)$ is asymptotically linear in $x$, that is: $$f_c(x) \sim kx$$ For some real constant $k \neq0$. I can do this by checking the derivative of $f$ with respect to $x$ (is there a better way?) However, I am not sure how to show $k$'s exact value, which is $-1/c$. I tried to use the Taylor expansion of $f_c(x)$ but the terms diverge; I have never studied asymptotic analysis before and am not sure how to proceed. Any guidance would be useful.

Answer 1

We can evaluate the limit $$\lim_{x\to\infty} \frac{c^x-(c+c^{-x}) ^x} {x} $$ directly.

The numerator can be written as $$-c^x((1+c^{-1-x}) ^{x} - 1)=-c^{x} \cdot\frac{t} {\log(1+t)}\cdot\log(1+t)$$ where $t\to 0$. Thus the desired limit equals the limit of $$-\frac{c^x\log(1+t)}{x}$$ ie $$-\frac{c^x\cdot x\log(1+c^{-1-x})}{x}$$ ie $$-c^ x\cdot c^{-1-x}\cdot\frac{\log(1+c^{-1-x})}{c^{-1-x}}$$ and the above tends to $-1/c$.

Answer 2

Since $(1+c^{-1-x})^x\approx 1+xc^{-1-x}$ for small $x$, $c^x[1-(1+c^{-1-x})^x]\approx-x/c$.