anonymous
  • anonymous
Evaluate limit (given in comment)
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[\lim_{x \rightarrow \infty} (\frac{ 11x-3 }{ 11x+5 })^{6x+1}\]
anonymous
  • anonymous
How do I check if it is indeterminate and how to begin evaluating it?
anonymous
  • anonymous
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 \[\exp\left(\lim_{x\to\infty}\frac{\ln\dfrac{11x-3}{11x+5}}{\dfrac{1}{6x+1}}\right)=\exp\left(\lim_{x\to\infty}\frac{88(6x+1)^2}{6(11x-3)(11x+5)}\right)=\cdots\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
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.
anonymous
  • anonymous
Thank you very much for the elaboration! I can solve the rest now.
anonymous
  • anonymous
You're welcome!

Looking for something else?

Not the answer you are looking for? Search for more explanations.