# Limit fraction

Gold Member
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?

Related Calculus and Beyond Homework Help News on Phys.org
malawi_glenn
Homework Helper
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 ?

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

Gold Member
so your expression goes as x^4/x = x^3 as x->oo, right?
Yup. But how come you factor x?

malawi_glenn
Homework Helper
what do you mean by "factor x"?

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

Gold Member
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
Homework Helper
Dearly Missed
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

I just note that x+(2/3)^x -> x as x->oo
What does that expression even mean??

malawi_glenn
Homework Helper
What does that expression even mean??
x + (2/3)^x has a linear asymptote with slope one as x -> oo

is that better?

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

The Bill
PeroK
Homework Helper
Gold Member
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}##.