Finding limit using L'Hospital rule

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Kenji Liew
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Hi, I am facing problem on solving this question.

Homework Statement



Given limx[tex]\rightarrow[/tex][tex]\infty[/tex](3x2+1)x

Homework Equations



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

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[tex]\rightarrow[/tex][tex]\infty[/tex] and ln (3x2+1) [tex]\rightarrow[/tex][tex]\infty[/tex], 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 +[tex]\infty[/tex]

Thanks!
 
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vela said:
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 said:
It might have been given to you just to see if you realize that the rule doesn't apply. :wink:
May be the lecturer try to test the students understanding.:smile: