0

Suppose you wish to calculate the following limit:

$$\lim_{x\to\infty}\frac{(1+x^2)^{-a}}{e^{-bx}}$$

for real numbers $a,b>0$. My instinct here is to rewrite as $$\lim_{x\to\infty}\left(\frac{1+x^2}{e^{\frac{bx}{a}}}\right)^{-a}$$ and then take the limit inside the parentheses: $$\left(\lim_{x\to\infty}\frac{1+x^2}{e^{\frac{bx}{a}}}\right)^{-a} \tag{$\ast$}$$ From here I would split the limit into two terms and use L’Hôpital’s rule on the second term to get the answer, $\infty$.

My question is whether the move I made in the starred equation is valid here, as I have only ever seen the power rule for limits stated for positive integer powers.

Rócherz
  • 3,976
wrb98
  • 1,295
  • 3
    This probably answers your question (basically the rule holds). – Sarvesh Ravichandran Iyer Dec 01 '23 at 04:19
  • (Limits and L’Hôpital’s rule are calculus, so why do you have this question tagged "precalculus"?) Limits may be exchanged with a function if the function is continuous at the limit of the inner expression: if $\lim_{x \to a} g(x) = b$ and $f(x)$ is continuous at $b$ (i.e., $\lim_{x \to b} f(x) = f(b)$, then $\lim_{x \to a} f(g(x)) = f(b)$. Alternatively, if there is some open interval about $a$ on which $g(x) \ne b$ (except possibly at $x = a$), then $\lim_{x \to a} f(g(x)) = \lim_{t \to b} f(t)$ even if $f$ is not continuous, or even not defined at $b$. – Paul Sinclair Dec 01 '23 at 22:37
  • Since $x^r = e^{r\ln x}$ for $x > 0$ and any $r$, and $e^x, \ln x,$ and $rx$ are continuous functions of $x$, this justifies the exchange of limit and exponent. – Paul Sinclair Dec 01 '23 at 22:43

0 Answers0