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My textbook says that this limit doesn't exist, but I don't understand - why? I tried calculating it by taking from both numerator and denominator factors that diverge to $\infty$ the fastest:

$$\lim_{n\to\infty}\frac{3^n+5^n}{(-2)^n+7^n}= \lim_{n\to\infty}\frac{5^n}{7^n}=0$$

Did I do something wrong here?

Integreek
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agromek
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3 Answers3

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Your result is correct but the way is not so much clear.

To solve properly we can observe that $\forall n\in\mathbb Z$:

$$\frac{3^n+5^n}{2^n+7^n}\le \frac{3^n+5^n}{(-2)^n+7^n}\le \frac{3^n+5^n}{-(2^n)+7^n}$$

and refer to squeeze theorem or more simply dividing by the by leading term $7^n$

$$\frac{3^n+5^n}{(-2)^n+7^n}=\frac{(3/7)^n+(5/7)^n}{(-2/7)^n+1}$$

Integreek
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user
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For an exponential function $f(x)=a^x$, the constraint is that $a\in R^+$. This is because for negative reals, the function is defined at certain discrete points only and is infinitely discontinuous, hence their graphs don't exist in the real $2D$ Cartesian Coordinate Plane.

Eg:- If we consider real numbers only, for $f(x)=(-2)^x$, $f\left(\frac1{2n}\right), n\in \mathbb Z$ does not exist. But $f\left(\frac1{2n+1}\right), n\in \mathbb Z$ exists. This means that there exist infinitely many points of discontinuity for $f(x)$.

So, in your question, the expression $(-2)^x$ is not well-defined over the set of reals. In fact, the question is wrong, unless there is some condition like $n=\frac{m}{2n+1} \text{ where } m,n\in \mathbb Z$ so that $(-2)^x$ is always well-defined. Or if instead of $-2$, there was some positive real say $\alpha$, then the limit would be zero:

P.S.:

Case 1: $0<\alpha<7$ $$\implies 7^n\text{ diverges faster than }\alpha^n\text{ as }n\to\infty $$ $$\implies\lim_{n\to\infty}\frac{3^n+5^n}{\alpha^n+7^n}= \lim_{n\to\infty}\frac{5^n}{7^n}=0$$ Case 2: $\alpha>7$ $$\implies \alpha^n\text{ diverges faster than }7^n\text{ as }n\to\infty $$ $$\implies\lim_{n\to\infty}\frac{3^n+5^n}{\alpha^n+7^n}= \lim_{n\to\infty}\frac{5^n}{\alpha^n}=0$$

Integreek
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  • Those who have doubts or require any clarification may kindly comment before downvoting. I don't see anything wrong with my answer. If there is, kindly comment. – Integreek Sep 15 '24 at 14:28
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    I didn’t downvote but maybe your discussion is not completely necessary to solve the limit. – user Sep 16 '24 at 05:38
  • Note also that the limit would exist also in the case that $x$ is real. This is an issue already discussed on MSE. I’ll give you some example. – user Sep 16 '24 at 05:46
  • Take a look here https://math.stackexchange.com/q/2889055 – user Sep 16 '24 at 05:48
  • @user , the link that you've referred to contains $\sin(\frac1{x})$, how does that relate to $(-2)^x$? – Integreek Sep 16 '24 at 06:42
  • According to the more general definition of limit we exclude from the domain the points where the expression is ill defined, therefore also for $(-2)^x$ it suffices that limit exists for $x\in \mathbb Z$. – user Sep 16 '24 at 06:59
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    Also:: it is usually assumed that when $n$ is used as a variable, it denotes an integer. And in that case, the problem discussed in this answer doesn´t show up. – Jan Olav R Sep 16 '24 at 07:37
  • @user okay I understood. But this does not apply to complex analysis, right? – Integreek Sep 16 '24 at 08:14
  • @JanOlavR I don't think so. In solutions of trigonometric equations, it is assumed that wherever $n$ is used, it is an integer but not in general situations like this one. – Integreek Sep 16 '24 at 08:14
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    Usually in calculus $n$ stands for an integer but it is a notation not a rule of course. Without any specification I always assume that it stands for an integer. – user Sep 16 '24 at 11:53
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    Yes of course I’m excluding complex analysis. – user Sep 16 '24 at 11:53
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In my opinion, the simplest way to solve this problem is to factor out $7^n$ from both numerator and denumerator, We get

$$ \lim_{n\to\infty}\frac{7^n}{7^n}\frac{(\frac{3^n}{7^n})+(\frac{5^n}{7^n})}{(\frac{(-2)^n}{7^n})+1}=\lim_{n\to\infty}\frac{(\frac{3^n}{7^n})+(\frac{5^n}{7^n})}{(\frac{(-2)^n}{7^n})+1}. $$

All the bases in the remaining powers in the quotient has absolute value less than one, so the limit of them is $0$.

Thus

$$ \lim_{n\to\infty}\frac{(\frac{3^n}{7^n})+(\frac{5^n}{7^n})}{(\frac{(-2)^n}{7^n})+1}=\frac{0+0}{0+1}=0. $$

As commented above, the statenent in the book is likely a misprint.

Jan Olav R
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