Solving an Infinite Limit: Should I Factor?

[Rectifier]

The problem
\lim_{x\rightarrow \infty} \frac{x^4 + x \ln x}{x + \left( \frac{2}{3} \right)^x}

The attempt
\lim_{x\rightarrow \infty} \frac{x^4 + x \ln x}{x + \left( \frac{2}{3} \right)^x} = \lim_{x\rightarrow \infty} \frac{x^4(1 + \frac{x \ln x}{x^4}) }{x + \left( \frac{2}{3} \right)^x}Should I factor $x$ or $\left( \frac{2}{3} \right)^x$ in the denomenator?

 

[malawi_glenn]

what will happen to x+ (2/3)^x as x-> oo ?

does x*ln(x) goes to oo faster or slower than x^4 as x-> oo ?

 

[Rectifier]

what will happen to x+ (2/3)^x as x-> oo ?

does x*ln(x) goes to oo faster or slower than x^4 as x-> oo ?

I am aware that $x^4$ approaches infinity faster than the other expression in numerator. I am only interested in the denominator at this point. As for the first question $x$
approaches infinity while $\left( \frac{2}{3} \right)^x$ approaches 0.

 

[malawi_glenn]

so your expression goes as x^4/x = x^3 as x->oo, right?

 

[Rectifier]

so your expression goes as x^4/x = x^3 as x->oo, right?

Yup. But how come you factor x?

 

[malawi_glenn]

what do you mean by "factor x"?

I just note that x+(2/3)^x -> x as x->oo

 

[Rectifier]

what do you mean by "factor x"?

I just note that x+(2/3)^x -> x as x->oo

I meant that you factorize x from denominator. So we get $\lim_{x\rightarrow \infty} \frac{x^4(1 + \frac{\ln x}{x^3}) }{x \left( 1 + \frac{ \frac{2}{3} ^x}{x} \right) } = \lim_{x\rightarrow \infty} \frac{x^3(1 + \frac{ \ln x}{x^3}) }{\left( 1 + \frac{ \frac{2}{3} ^x}{x} \right) }$.

 

[Ray Vickson]

I meant that you factorize x from denominator. So we get $\lim_{x\rightarrow \infty} \frac{x^4(1 + \frac{\ln x}{x^3}) }{x \left( 1 + \frac{ \frac{2}{3} ^x}{x} \right) } = \lim_{x\rightarrow \infty} \frac{x^3(1 + \frac{ \ln x}{x^3}) }{\left( 1 + \frac{ \frac{2}{3} ^x}{x} \right) }$.

That would be one way to do it; another way would be to use l'Hospital's Rule; see, eg.,
http://tutorial.math.lamar.edu/Classes/CalcI/LHospitalsRule.aspx

 

[micromass]

I just note that x+(2/3)^x -> x as x->oo

What does that expression even mean??

 

[malawi_glenn]

What does that expression even mean??

x + (2/3)^x has a linear asymptote with slope one as x -> oo

is that better?

 

[micromass]

x + (2/3)^x has a linear asymptote with slope one as x -> oo

is that better?

So basically, you mean $\lim_{x\rightarrow +\infty}\frac{x + (2/3)^x}{x} = 1$. The mathematical notation for that is $x + (2/3)^x \sim x$ https://en.wikipedia.org/wiki/Big_O_notation#Family_of_Bachmann.E2.80.93Landau_notations

 

[malawi_glenn]

Wasn't it obvious what I was trying to communicate?

 

[PeroK]

The problem
\lim_{x\rightarrow \infty} \frac{x^4 + x \ln x}{x + \left( \frac{2}{3} \right)^x}

The attempt
\lim_{x\rightarrow \infty} \frac{x^4 + x \ln x}{x + \left( \frac{2}{3} \right)^x} = \lim_{x\rightarrow \infty} \frac{x^4(1 + \frac{x \ln x}{x^4}) }{x + \left( \frac{2}{3} \right)^x}Should I factor $x$ or $\left( \frac{2}{3} \right)^x$ in the denomenator?

Another technique is to estimate the numerator and/or denominator. In this case the numerator is greater than $x^4$ and the denominator is less than $2x$, so the overall expression is greater than $\frac{x^4}{2x}$.