I know that when we use the squeeze (or sandwich) theorem in these kinds of questions where exponential functions are involved, we prioritize the variable with the highest value, which in this case, is $9^x$, and use it in upper/lower bounds. Though the main issue for me here is the degree of the root ($2x+1$), I have only dealt with roots that are either the same or very similar with the exponential function itself, like this: $\sqrt[x]{3^x+5^x}$, where the root is pretty much the same as the power of $3$ and $5$; but here, it is quite different than the power of exponential.

I (as an amateur in Squeeze Theorem) didn't know what to use to deal with that root. What would you recommend to put in upper/lower bounds and why?

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    $\begingroup$ The reason that you maintain the highest-value variable is because it 'drives the value'; $4^x$ is much smaller than $9^x$, and that can be formalized. In this case, I would sandwich the inside between $0^x+9^x$ and $9^x+9^x$; both can be simplified in ways that should let you compute the respective limits. The fact that the root is a slightly more complicated function is largely moot to the procedure; in this case, you'll want to use something like the fact that $9^x=3^{2x+1}/3$ (why?)... $\endgroup$ – Steven Stadnicki Apr 8 at 23:05

Since $9^x<4^x+9^x<2\,\cdot 9^x$

$$9^{x/(2x+1)}\leq \big(4^x+9^x)^{1/(2x+1)}\leq 2^{1/(2x+1)}9^{x/(2x+1)}$$

On the left-end of the inequalities one gets convergence to $9^{1/2}$; on the right-end one gets $1\cdot 9^{1/2}=9^{1/2}$

  • $\begingroup$ Hey, thank you. You gave me a good idea. $\endgroup$ – 5Nik Apr 8 at 23:50
  • $\begingroup$ @Laz, what does he need to fix? $\endgroup$ – 5Nik Apr 8 at 23:51

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