Why is $\lim\limits_{x \to 0^+}(1+ x)^\frac1x$ is $e$ and not $\infty$

calculuslimits

I am a grade 12th student. I am having doubts about limits as follows:-

If $a>1$, then $\lim\limits_{x \to \infty}a^x = \infty$.

The above is because a number greater than $1$ when multiplied with itself, increases. If we do it a large number of times, the number will get close to $\infty$. Therefore its limiting value becomes $\infty$.

Applying same logic $\lim\limits_{x \to 0^+}(1+ x)^\frac1x = \infty$.

But $\lim\limits_{x \to 0^+}(1+ x)^\frac1x = e$.

Can anyone suggest to me where I am wrong?

Edit

I had asked my mentor about this. He said the following.

In case of $\lim\limits_{x \to \infty}a^x$, base and exponent are independent of each other whereas in case of $\lim\limits_{x \to 0^+}(1+ x)^\frac1x$, the base and exponent are dependent on each other. Because of this dependence, as soon as we create the base, i.e. $(1+x)$, $x$ becomes fixed and as a result the exponent becomes fixed and can't approach infinity. However, for $a^x$, $x$ can become as small as we want and therefore tends to infinity.

Can anyone please explain me why are we trying to fix the base in case of $\lim\limits_{x \to 0^+}(1+ x)^\frac1x$? Is there any such rule that we have to first create a base and then apply the limits? Am I having a wrong interpretation of limits? Can anyone please interpret his statements in a simplified language.

Any help is appreciated. Thank you.

Best Answer

The issue is that the first statement should really be

If $a$ is a constant greater than $1$, then $\lim_{x\to\infty}a^x=\infty$.

It is normally understood that if you say "Let $a>1$", or similar, that you are talking about a specific, constant real number $a$.

Here, it's important that $a$ does not depend on $x$, and without that restriction the statement wouldn't be true. It's easier to see this by replacing $a$ with $\sqrt[x]2$; then certainly $\sqrt[x]2>1$ for any $x$, but $(\sqrt[x]2)^x=2$ for every $x$, so its limit is also $2$.

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