Limits approaching infinity with x in the exponent

[bschiff10]

What is the limiting value as x approaches infinity of ((x^9 + 4 E^(0.6 x))/(3 x^12 + 2 E^(0.6 x)))

I can't figure out which is the dominant term. The section with e raised to a power with x in it, if substituted, will equal infinity. Then we would be left with x^9/3x^12. So would the answer be 1/3x^3?

Also, I have no idea about where to start with this one.

Find the limiting value as x approaches infinity of (3 x^8 - 123 Cos[x] - 6 x^2)/E^(0.4x).

Thank you so much,
Briana

 

[BigGlenntheHeavy]

\(\displaystyle First \ one: \ f(x) \ = \ \lim_{x\to \infty}\frac{x^{9}+4e^{.6x}}{3x^{12}+2e^{.6x}}\)

\(\displaystyle Now, \ f(x) \ gives \ the \ indeterminate \ form \ \frac{\infty+\infty}{\infty+\infty}, which \ is \ \frac{\infty}{\infty}\)

\(\displaystyle Ergo, \ the \ Marqui \ to \ the \ rescue, \ to \ wit: \ \lim_{x\to\infty}\frac{0+4Ke^{.6x}}{0+2Ke^{.6x}} \ = \ 2.\)

 

[soroban]

Hello, bschiff10!

\(\displaystyle \lim_{x\to\infty}\frac{x^9 + 4e^{0.6 x}}{3 x^{12} + 2e^{0.6 x}}\)

The exponential function is dominant.


\(\displaystyle \text{Divide top and bottom by }e^{0.6x}\)

. . \(\displaystyle \lim_{x\to\infty}\,\frac{\dfrac{x^9}{e^{0.6x}} + 4}{\dfrac{3x^{12}}{e^{0.6x}} + 2} \;=\;\frac{0+4}{0+2} \;=\;2\)

 

[bschiff10]

Thank you so much!

 

[galactus]

\(\displaystyle \lim_{x\to \infty}\frac{3 x^{8} - 123 Cos[x] - 6 x^{2}}{e^{0.4x}}\)

Kind of a strange limit problem.

As in the first one, the e term is dominant.

As it gets larger and larger, what does the limit approach?.

Take each separate and they all approach the same thing.

\(\displaystyle 3\lim_{x\to \infty}\frac{x^{8}}{e^{\frac{2x}{5}}}\)

\(\displaystyle 123\lim_{x\to \infty}\frac{cos(x)}{e^{\frac{2x}{5}}}\)

\(\displaystyle 6\lim_{x\to \infty}\frac{x^{2}}{e^{\frac{2x}{5}}}\)

 

[chrisr]

e[sup:2416cwbv]x[/sup:2416cwbv] can be written as 1+x+x[sup:2416cwbv]2[/sup:2416cwbv]/2!+x[sup:2416cwbv]3[/sup:2416cwbv]/3!+x[sup:2416cwbv]4[/sup:2416cwbv]/4!+......... an infinite series.

When x is 10!, the (10!)[sup:2416cwbv]10[/sup:2416cwbv]/10! is (10!)[sup:2416cwbv]9[/sup:2416cwbv] which is x[sup:2416cwbv]9[/sup:2416cwbv],
hence e[sup:2416cwbv]x[/sup:2416cwbv] is already dwarfing x[sup:2416cwbv]9[/sup:2416cwbv].

Similarly the x[sup:2416cwbv]12[/sup:2416cwbv] will be only a single term of e[sup:2416cwbv]x[/sup:2416cwbv] for a certain x as x increases.

e[sup:2416cwbv]0.6x[/sup:2416cwbv] is greater than sqrt(e[sup:2416cwbv]x[/sup:2416cwbv]),
but if we expand the powers, we only need go to the term in e[sup:2416cwbv]x[/sup:2416cwbv] with twice the power to equate
the x[sup:2416cwbv]9[/sup:2416cwbv] and x[sup:2416cwbv]12[/sup:2416cwbv] to a single term of e[sup:2416cwbv]0.5x[/sup:2416cwbv].

Thus, e[sup:2416cwbv]0.6x[/sup:2416cwbv] dwarfs x[sup:2416cwbv]9[/sup:2416cwbv] and x[sup:2416cwbv]12[/sup:2416cwbv] as x approaches infinity.

As e is approximately 2.7, it would be a different story to compare it with k[sup:2416cwbv]x[/sup:2416cwbv].