Is $2^{\infty}$ an Indeterminate form

Question

We know that when $$\lim_{x \to a}f(x) \to 1$$ and $$\lim_{x \to a}g(x) \to \infty$$ then $$\lim_{x\to a}f(x)^{g(x)}$$ is an indeterminate form since in the neighbourhood of $a$ we cannot predict the exact value to where the limit approaches.

Is it indeterminate if the limit is in $2^{\infty}$ form?

Answer 1

No , $2^\infty$ is not an indeterminate form.

If $\displaystyle\lim_{x\to a}f(x)=2$ and $\displaystyle\lim_{x\to a}g(x)=\infty$ then to get $f(x)^{g(x)}>M$, just make sure $x$ is close enough to $a$ so that $f(x)>1.1$ for any closer value (possible since $f(x)$ can be kept close to $2$) and so that $g(x)$ is greater than $\log_{1.1}M$ (possible since $g(x)$ can be made arbitrarily large).

Answer 2

Rewrite $2^x$ as $e^{x\ln 2}$. You have

• $e^u \xrightarrow[u\to \infty]{} \infty$;

• The function $\exp$ is continuous; and

• $x\ln 2 \xrightarrow[x\to \infty]{} \infty$ (as $\ln 2 > 0$).

Combining the 3, $e^{x\ln 2} \xrightarrow[x\to \infty]{} \infty$.

Not that this does not work for your case $f(x) \xrightarrow[x\to \infty]{} 1$, as then $\ln f(x) \xrightarrow[x\to \infty]{} 0$ and you do not have the argument "$x\ln f(x) \xrightarrow[x\to \infty]{} \infty$." (There is an indeterminate form there.)