Finding limit using L'Hospital rule

[Kenji Liew]

Hi, I am facing problem on solving this question.

Homework Statement

Given lim_{x\rightarrow\infty}(3x²+1)^x

Homework Equations

From the above limit, if we substitute x\rightarrow\infty, obviously we don't have the indeterminate forms: \infty^0 , 0⁰ or 1^\infty.

The Attempt at a Solution

However, I try to attempt this question by putting ln for both sides and the equation become ln y = x ln (3x²+1). I get x\rightarrow\infty and ln (3x²+1) \rightarrow\infty, this indeterminate form does not exist. I wonder how to solve using L'Hospital rule? If I plotting the graph using mathematica, it gives the answer of limit that is +\infty

Thanks!

 

[vela]

The Hospital rule doesn't apply because, as you noted, you don't get any of the indeterminate forms. In other words, there's no question the function blows up as x goes to infinity.

 

[Kenji Liew]

The Hospital rule doesn't apply because, as you noted, you don't get any of the indeterminate forms. In other words, there's no question the function blows up as x goes to infinity.

Thanks for replying.I think something wrong with this question.

 

[vela]

It might have been given to you just to see if you realize that the rule doesn't apply.

 

[Kenji Liew]

It might have been given to you just to see if you realize that the rule doesn't apply.

May be the lecturer try to test the students understanding.