how to solve lim x->+infinity for [x+e^x+e^(2x) ]^(1/x). I need detail solution.Anyone can help?

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.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

how to solve lim x->+infinity for [x+e^x+e^(2x) ]^(1/x). I need detail solution.Anyone can help?

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

Trying to solve this by "plugging in infinity" will give us an indeterminate result (infinity raised to 0), so we have to manipulate this to another form that can give an answer. Usually, an exponent of anything else besides an integer calls for taking the natural log of the limit argument, and then raising e to that result (because taking natural log and exponentiating cancels out).
So, \[\lim_{x \rightarrow \infty} (x + e^x + e^{2x})^{1/x}\] \[= e ^{\lim_{x \rightarrow \infty} \ln((x+e^x+e^{2x})^{1/x}}\]
A property logarithms allows us to move the (1/x) to the front of the natural log, so:

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

\[e ^{\lim_{x \rightarrow \infty} \frac{1}{x}\ln(x + e^x + e^{2x})}\]
at this point, trying to evaluate the limit will give us infinity over infinity, but at least it's in a form that allows us to use another technique, L'hopital's rule, which involves taking the limit of the derivative of the numerator, over the derivative of the denominator
\[e ^{\lim_{x \rightarrow \infty} \frac{\frac{1+e^x + 2e^{2x}}{x+e^x + e^{2x}}}{1}}\]
If you keep doing L'hopital's rule, it would never end, but it suffices to recognize as x grows, only the leading term has an affect on the value, so we can take the limit of [2e^(2x)][e^(2x)], which gives 2. Finally, exponentiate to get the aswer: e^2
not really understand this part - take the limit of [2e^(2x)][e^(2x)], which gives 2.
Oops, typo. It should read "take the limit of [2e^(2x)]/[e^(2x)], which gives 2". Does that make it clearer?
Thank you! =)

Not the answer you are looking for?

Search for more explanations.

Ask your own question