# When can you treat a limit like an equation?

Lately, I've been very confused about the weird properties of limits. For example, I was very surprised to find out that $\lim_{n \to \infty} (3^n+4^n)^{\large \frac 1n}=4$ , because if you treat this as an equation, you can raise both sides to the $n$ power, subtract, and reach the wrong conclusion that $\lim_{n \to \infty} 3^n=0$ . I've asked this question before over here, and the answer was that $\infty-\infty$ is not well defined. I also found out here that you cannot raise both sides of a limit to a power unless the limit is strictly less than $1$ . However, there are also many examples where limits are treated as equations. For example, taking the logarithm of each side is standard procedure. Substitutions such as using $\lim_{x \to 0} \frac {\sin x}{x}=1$ work (although other substitutions sometimes don't work). So when can a limit be treated as an equation? Can you take for example the sine or tangent of each side like you can take the log? My guess is that you can treat it as an equation $at$ $least$ whenever $nothing$ is approaching $0$ or $\infty$ , but I'm not sure. Thanks.

P.S. Please keep the answers at a Calculus 1 level, and I have not learned the epsilon-delta definition of a limit.

• WHat do you mean by «treat this as an equation»? – Mariano Suárez-Álvarez Jul 11 '13 at 3:09
• Limit yourself to operating with limits according to the rules that the theorems you know about limits allow. – Mariano Suárez-Álvarez Jul 11 '13 at 3:11
• @Mariano: I’m pretty sure that Ovi is talking about moving operations through the limit, e.g., replacing $$\left(\lim_{n\to\infty}(3^n+4^n)^{1/n}\right)^n$$ by $$\lim_{n\to\infty}\left((3^n+4^n)^{1/n}\right)^n\;.$$ – Brian M. Scott Jul 11 '13 at 3:15
• @MarianoSuárez-Alvarez: "Limit yourself to operating with limits according to the rules that the theorems you know about limits allow." — I think the OP is asking exactly what all the rules are that the theorems about limits allow. :-) Of course an exhaustive list cannot be given as an answer, but something useful could still be said, I think... – ShreevatsaR Jul 11 '13 at 3:22
• The problem here has not much to do with limits, really. For example, $(\int_0^1x\,\mathrm dx)^2$ is not the same as $\int_0^1x^2\,\mathrm dx$, or $1+\sqrt 2\neq\sqrt{1+2}$. If you want, my suggestion is for the OP to reflect about what putative rule he is wanting to use. – Mariano Suárez-Álvarez Jul 11 '13 at 3:30

The $n$ in the limit has no meaning outside of the limit. Therefore you cannot raise both sides to the $n$th power and then "bring in" the $n$ from the outside into the limit (to be combined with the $1/n$ exponent, as you seem to be doing). In an expression such as $$\lim_{n \to \infty} (3^n+4^n)^{1/n}$$ $n$ is a "dummy" variable. It simply tells you which variable in the inner expression we are taking the limit with respect to. The value of the limit is not a function of $n$, and therefore "raising both sides to the $n$th power" is not a meaningful operation.

It is, however, okay to square both sides (or cube, or raise to any fixed power) of the equation, or take the log, sin, tan, or any other function. An equation involving a limit is still an equation and you can always do any operation to both sides of an equation. However, you still have to be careful. Suppose you have an equation like $$\lim_{x \to 0} f(x) = L.$$ Now you take the log of both sides. You get $$\log \lim_{x \to 0} f(x) = \log L.$$ Notice I have not yet brought the $\log$ "into" the limit, which is usually the next step you want to take. In order for that step to be valid, you need to know that $\log$ is continuous at the values of $f(x)$ near $x=0$, and then you would get $$\lim_{x \to 0} \log f(x) = \log L.$$

• So you could also take in any trigonometric function that is continuous around whatever your variable is approaching? (for example 0 in your example). Also if you check the second link I provided, according to the accepted answer you apparently $can$ bring in the power iff the limit is less that $1$ so I don't see how your argument about exponents holds in that case . Also, I never claimed that "the value of a limit is a function of n", and I don't see how I ever implied it. – Ovi Jul 11 '13 at 3:54
• The justification for the process in the other answer is not "raise both sides to the $g(x)$ power". That is meaningless as I explained above. Instead, you start with $\lim_{x \to 0} f(x)^{g(x)}$ and then rewrite this as $$(\lim_{x \to 0} f(x))^{\lim_{x \to 0} g(x)}$$ to get what you want. However, this step is not always valid and is subject to some conditions, as explained in the other answer. (It is similar to saying that the limit of a sum (or product) is the sum (or product) of limits, only with powers. But you have to be more careful with powers.) – Ted Jul 11 '13 at 4:16
• To take a simpler version of the above: Suppose we are trying to calculate $\lim_{x \to 1} x^2 + x$. You wouldn't do this by calculating $\lim_{x \to 1} x^2 = 1$ and then "adding $x$ to both sides". – Ted Jul 11 '13 at 4:19
• @Ovi Note that in the other answer, the author was probably thinking of the case $\lim_{x \to c} g(x) = \infty$ when they wrote that the limit of $f(x)$ has to be less than 1. If $\lim_{x \to c} g(x)$ is positive and finite then it doesn't matter what the limit of $f(x)$ is. – Ted Jul 11 '13 at 4:46
• So this means that if the exponent is not approaching infinity (or probably 0 as well) I can always raise each power to that exponent and bring it into the limit? – Ovi Jul 11 '13 at 5:26

This is far from a complete answer, but I will try to address what I think might be the main problem.

If the function $f$ is continuous at the limit, then we have $$\lim_{n\to\infty}f\left(\left(3^n+4^n\right)^{1/n}\right)=f(4)\tag{1}$$ That is we can pass $f$ across the limit. However, since $n$ varies as part of the limit (that is, it is bound to the limit as a dummy variable), we cannot use $f(x)=x^n$ to get $$\lim_{n\to\infty}3^n+4^n=4^n\tag{2}$$ The $n$ in $\lim\limits_{n\to\infty}\left(3^n+4^n\right)^{1/n}$ is only meaningful inside the limit. If we write $\left(\lim\limits_{\color{#00A000}{n}\to\infty}\left(3^{\color{#00A000}{n}}+4^{\color{#00A000}{n}}\right)^{1/\color{#00A000}{n}}\right)^\color{#C00000}{n}$, the green $n$ tends to $\infty$ whereas the red $n$ does not. The green $n$ is a dummy variable and could just as easily be replaced by any other variable, and it has no fixed value. The red $n$ affects the value of the expression, and it needs a value for the expression to have a value.

I see that in this answer, the validity of $$\lim_{n\to\infty}f_n(x_n)=\lim_{k\to\infty}f_k(\lim_{n\to\infty}x_n)\tag{3}$$ is considered. If $f_k\to f$ uniformly in a neighborhood of $\lim\limits_{n\to\infty}x_n$, and $f$ is continuous at $\lim\limits_{n\to\infty}x_n$, then $(3)$ is valid.

Since $f_k(x)=x^k$ converges uniformly to $0$ on $[-1+\epsilon,1-\epsilon]$ for $\epsilon>0$, if $\lim\limits_{n\to\infty}x_n\in(-1,1)$, then both sides of $(3)$ tend to $0$.

However, it should be understood that there is absolutely no connection between the index of the outer limit and the index of the inner limit on the right side of $(3)$. In fact for each $k$, $n\to\infty$.

If we let $f_k(x)=x^k$ and $x_n=(3^n+4^n)^{1/n}$, $(3)$ becomes $$\lim_{n\to\infty}\left((3^n+4^n)^{1/n}\right)^n=\lim_{k\to\infty}\left(\lim_{n\to\infty}(3^n+4^n)^{1/n}\right)^k\tag{4}$$ However, since $\lim\limits_{n\to\infty}(3^n+4^n)^{1/n}=4$ and $f_k$ does not converge uniformly in a neighborhood of $4$, $(4)$ does not hold, and we cannot assume $k=n$ or think of the right side as $\lim\limits_{n\to\infty}3^n+4^n$, especially since that limit is not finite.

• The main problem I have is when you said "However, since n varies as part of the limit, we cannot use..." . Can you further explain why "n varying as part of the limit", you cannot raise the limit to the $nth$ power? Also, if you look at the second link I put up the accepted answer says you $can$ raise each side to a power of $n$ iff the limit is less than $1$. – Ovi Jul 11 '13 at 3:57