Kenji Liew
- 25
- 0
Hi, I am facing problem on solving this question.
Given limx[tex]\rightarrow[/tex][tex]\infty[/tex](3x2+1)x
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].
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!
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!
Last edited: