Question

Evaluate limit (given in comment)

Answer 1

\[\lim_{x \rightarrow \infty} (\frac{ 11x-3 }{ 11x+5 })^{6x+1}\]

Answer 2

How do I check if it is indeterminate and how to begin evaluating it?

Answer 3

When you directly "substitute" \(x=\infty\), you get a nasty indeterminate form \(\left(\dfrac{\infty}{\infty}\right)^\infty\). There a few methods to work around this.

\(\textbf{Method 1:}\) rearrange the limit to make it an appropriate indeterminate form in order to apply L'Hopital's rule. In particular,
\[\left(\frac{11x-3}{11x+5}\right)^{6x+1}=\exp\left(\ln\left(\frac{11x-3}{11x+5}\right)^{6x+1}\right)=\exp\left((6x+1)\ln\left(\frac{11x-3}{11x+5}\right)\right)\]Since \(\exp x\) is continuous, you can rewrite the limit as
\[\lim_{x\to\infty}\exp\left((6x+1)\ln\frac{11x-3}{11x+5}\right)=\exp\left(\lim_{x\to\infty}(6x+1)\ln\frac{11x-3}{11x+5}\right)\]Evaluating directly again gives another not-so-useful indeterminate form \(\infty\times0\), since \(6x+1\to\infty\) and \(\ln\frac{11x-3}{11+5}\to0\). To remedy this, put the first factor in the denominator, like so:
\[(6x+1)\ln\frac{11x-3}{11x+5}=\frac{\ln\dfrac{11x-3}{11x+5}}{\dfrac{1}{6x+1}}\]which yields the desirable \(\dfrac{0}{0}\) indeterminate form. No you can apply L'Hopitals' rule. You have
\[\exp\left(\lim_{x\to\infty}\frac{\ln\dfrac{11x-3}{11x+5}}{\dfrac{1}{6x+1}}\right)=\exp\left(\lim_{x\to\infty}\frac{88(6x+1)^2}{6(11x-3)(11x+5)}\right)=\cdots\]

Answer 4

Typo: I'm missing a negative sign in front of the limit in the last line on the RHS.

Other methods to briefly mention...

\(\textbf{Method 2:}\) recalling the definition of the exponential function, you have
\[e^t=\lim_{x\to\infty}\left(1+\frac{t}{x}\right)^x=\lim_{x\to\infty}\left(\frac{x+t}{x}\right)^x\]The idea would be the rearrange \(\left(\dfrac{11x-3}{11x+5}\right)^{6x+1}\) to get similar form, then it's just a matter of recognizing which term is \(t\).

\(\textbf{Method 3:}\) you can try using the series expansion of the function.

Answer 5

Thank you very much for the elaboration! I can solve the rest now.

Answer 6

You're welcome!