Squeezing The Log

Calculus Level 2

$\large \displaystyle \lim_{x \to \infty} \dfrac{\log{(x^3+(\log{(x)})^3)}}{\log{(x^2+(\log{(x)})^2)}} = \, ?$

Clarification : The base of all the logarithms are equal but not specified.

0

\frac{3}{4}

\frac{3}{2}

3

2 solutions

Jack Lam
Apr 29, 2016

Relevant wiki: Squeeze Theorem

For sufficiently large $x$ , the following are true:

$x^3 < x^3 + \log^3{(x)} < 2x^3$

$x^2 < x^2 + \log^2{(x)} < 2x^2$

Take logarithms of all three sides of both inequalities, which is valid, as the logarithm preserves order over the positive reals.

$3\log{x} < \log{(x^3 + \log^3{(x)})} < 3\log{x} + \log{2}$

$2\log{x} < \log{(x^2 + \log^2{(x)})} < 2\log{x} + \log{2}$

$\Rightarrow \frac{3\log{x}}{2\log{x}+\log{2}} < \frac{\log{(x^3 + \log^3{(x)})}}{\log{(x^2 + \log^2{(x)})}} < \frac{3\log{x}+\log{2}}{2\log{x}}$

Dividing throughout the upper and lower bounds, taking the limit and applying the squeeze theorem is sufficient to show the value of the limit, all of which is left as an exercise to the reader.

log a/log b =log (a-b) >>>>>>>put a- p=10^y, when x> infinitely then with Divisible / x^3 the right of equal =1(thats is rong may be but i am answer quickly) then y =0

Patience Patience - 5 years, 1 month ago

Your very first statement is not a rule. The quotient of two Logarithms is a meaningless statement. The quotient of two powers with identical bases is what you are thinking of.

Jack Lam - 5 years, 1 month ago

you are true ,sorry, but i am will take time thats nice questions (not specified)

Patience Patience - 5 years, 1 month ago

Vishesh Agarwal
Dec 12, 2018

Use L hospital rule and then set try to set all logs as logx /x and then as X tend to infinity logx/X goes to 0

Squeezing The Log

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