Monday, 27 May 2019

calculus - Limit of sqrt4x2+3x2x as xtoinfty



limx4x2+3x2x



I thought I could multiply both numerator and denominator by 1x, giving



limx4+3x21x




then as x approaches infinity, 3x essentially becomes zero, so we're left with 2-2 in the numerator and 1x in the denominator, which I thought would mean that the limit is zero.



That's apparently wrong and I understand (algebraically) how to solve the problem using the conjugate, but I don't understand what's wrong about the method I tried to use.


Answer



Hint: 4x2+3x2x=3x4x2+3x+2x=34+3x+2


No comments:

Post a Comment

real analysis - How to find limhrightarrow0fracsin(ha)h

How to find limh0sin(ha)h without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...