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I am having some trouble calculating limits. I looked it up and I cannot seem to find a definite method for calculating limits, all I seem to find are methods that find the solution through logic or even by guessing.

I can solve some limits by using logic or/and by looking at the graph of the function, but I can only do that for the easiest/most obvious ones.

Could you solve these examples for me and explain to me how you did it step by step, preferably using a more analytical method that could be applied to the majority of, if not all, similar problems?

Examples:

$$\lim_{x\to+\infty}(\frac{5^x}{3^x})$$ $$\lim_{x\to+\infty}(\frac{3^{x+1}}{\pi^x})$$ $$\lim_{x\to+\infty}(\frac{100^x}{x^{100}})$$ $$\lim_{x\to+\infty}(2^{-x}\cdot x^4)$$ $$\lim_{x\to-\infty}(3^{-x})$$ $$\lim_{x\to-\infty}(\frac{3^{-x}}{x^4})$$

Mark Read
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2 Answers2

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For the first and second ones, you can rewrite the functions as $(5/3)^x$ and $3\cdot(3/\pi)^x$. Since $5/3>1$, the first limit is $\infty$. Since $0<3/\pi<1$, the second limit will be $0$.

The fifth one is just the limit as $x\to\infty$ of $3^x$ which is $\infty$.

For the other ones, you could use the fact that exponential growth is always faster than polynomial growth. See anon's answer here for example.

Community
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Guest
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A lot of the time, the way to compute these limits is to simply consider the behavior of these functions as $x$ gets bigger and bigger. For example:

$$ \lim_{x\rightarrow +\infty} (\cfrac{5^x}{3^x}) $$

In your mind, fix $x$ to be some reasonable number, say $2$. The numerator then is $5^2 = 25$, while the denominator is $3^2 = 9$. Now consider a larger $x$, say $3$. The numerator is $5^3 = 125$, while the denominator is $3^3 = 81$. In terms of growth, we can see that increasing $x$ by one will increase the numerator by $5$ and the denominator by $3$.

That's all good and fine, but how does this relate to taking the limit? Well now consider increasing $x$ by a lot. Think finite amounts first, say $100$. Then the overall quantity has increased by $(\frac{5}{3})^{100} \approx 10^{22}$ - a huge ratio! I hope you can now convince yourself that going to infinity, we see that this ratio will only increase, so the $\lim_{x\rightarrow +\infty}$ does not exist (or is infinite).

One more example: $$\lim_{x\rightarrow +\infty}2^{-x} x^4 = \lim_{x\rightarrow +\infty} \cfrac{x^4}{2^x}$$

Again, we consider a reasonable number to test $x$ for: say $x=2$. The numerator is $2^4 = 16$, while the denominator is $2^2 = 4$. Considering a larger value such as $x = 4$, the numerator is $4^4 = 256$ while the denominator has become $2^4 = 16$. In terms of growth, increasing $x=2$ to $x=4$, the numerator scaled faster than the denominator for this choice of $x$.

You might think this means that the limit will again go to $\infty$ as $x\rightarrow\infty$, but we must be careful with our analysis. Looking closer, we see that we can compare the scaling of both the numerator and denominator from $x$ to $2x$:

$$numerator(2x) = (2x)^4 = 2^4(x^4) = 16 (numerator(x))$$

$$denominator(2x) = 2^{2x} = (2^{x})^2 = (denominator(x))^2$$

So doubling our choice of $x$ multiplies the numerator by $16$ and squares our denominator. For large values of $x$ (specifically $x>16$), I think you can convince yourself that squaring a number scales much faster than multiplying by $16$.

So the denominator will completely dominate the numerator as $x$ gets larger and larger. Therefore, we actually have that this ratio will only decrease as $x\rightarrow\infty$, and indeed $\lim_{x\rightarrow +\infty} 2^{-x}x^4 = 0$.

You asked for an analytical method to solve these problems. There isn't one. Rather, you need to carefully consider the asymptotic behavior of your functions as $x\rightarrow\infty$. Once you have considered the behavior of $a^x$, $x^b$, $\sin{x}$, $\log{x}$, and a few other common functions in isolation, you'll be able to piece together the behavior of most of these problems by looking at them in pieces.

Cheers. I hope this helps.

NoseKnowsAll
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