# Math Help - two ln problems

1. ## two ln problems

1. The rate of transmission in a telegraph cable is observed to be proportional to x^2\ln (1/x), where x is the ratio of the radius of the core to the thickness of the insulation ( 0<x<1 ). What value of x gives the maximum rate of transmission?
I overlooked this simplicity of this problem. I just needed to find the deriv of the problem where it = 0. This was .60653019.

2. f(x)=ln(x)/(1+ln(x)^2). Find lim as x->0+ and x-> inf
nevermind, it's 0 for both limits

2. Originally Posted by thedoge
1. The rate of transmission in a telegraph cable is observed to be proportional to x^2\ln (1/x), where x is the ratio of the radius of the core to the thickness of the insulation ( 0<x<1 ). What value of x gives the maximum rate of transmission?...
Hello,

I presume the function is:

$t(x)=x^2 \cdot \ln(x)$. You are looking for a maximum of t(x). This will happen if t'(x) = 0. Thus calculate the first derivative. Use the product rule:

$t'(x) = \ln(x) \cdot 2x + x^2 \cdot \frac{1}{x}=x(2 \ln(x) + 1)$

Now t'(x) = 0. Solve for x:

$x(2 \ln(x) + 1)=0 \Longleftrightarrow \underbrace{x = 0}_{\text{not allowed here}} \text{ or } \ln(x)=- \frac{1}{2}$

$\ln(x)=- \frac{1}{2} \Longleftrightarrow x=\sqrt{\frac{1}{e}} \approx 0.606530659...$

EB

3. Originally Posted by thedoge
...

2. f(x)=ln(x)/(1+ln(x)^2). Find lim as x->0+ and x-> inf
nevermind, it's 0 for both limits
Hello,

you are supposed to know

(A) ---> $\lim_{x \rightarrow \infty}{(\ln(x))}=\infty$

and because $\lim_{y \rightarrow \infty}{\frac{1}{y}}=0$

(B) ---> $\lim_{x \rightarrow 0}{(\ln(x))}=\lim_{y \rightarrow \infty}{\left( \ln \left( \frac{1}{y} \right) \right)}=\lim_{y \rightarrow \infty}{(-\ln(y))}=-\infty$

$f(x)=\frac{\ln(x)}{1+(\ln(x))^2}$

$\lim_{x \rightarrow \infty}{f(x)}$ is undefined. Therefore use the rule de l'Hôpital:

$\lim_{x \rightarrow \infty}{\frac{\ln(x)}{1+(\ln(x))^2}}=\lim_{x \rightarrow \infty}{\frac{\frac{1}{x}}{2\ln(x) \cdot \frac{1}{x}}}=\lim_{x \rightarrow \infty}{\frac{1}{2\ln(x) \cdot }}=0$

According to (A) the limit is zero because the denominator approaches infinity.

$\lim_{x \rightarrow 0}{\frac{\ln(x)}{1+(\ln(x))^2}}=\lim_{y \rightarrow \infty}{\frac{\frac{1}{\frac{1}{y}}}{2\ln(\frac{1} {y}) \cdot \frac{1}{\frac{1}{y}}}}=\lim_{y \rightarrow \infty}{\frac{1}{2\ln(\frac{1}{y})}}=\lim_{y \rightarrow \infty}{\frac{1}{-2\ln(y) }} = 0$

EB