Monday, 27 May 2019

calculus - Limit of sqrt4x2+3x2x as xtoinfty



lim



I thought I could multiply both numerator and denominator by \frac{1}{x}, giving



\lim_{x\to\infty}\frac{\sqrt{4 + \frac{3}{x}} -2}{\frac{1}{x}}




then as x approaches infinity, \frac{3}{x} essentially becomes zero, so we're left with 2-2 in the numerator and \frac{1}{x} 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: \sqrt{4x^{2}+3x}-2x=\frac{3x}{\sqrt{4x^{2}+3x}+2x}=\frac{3}{\sqrt{4+\frac{3}{x}}+2}


No comments:

Post a Comment

real analysis - How to find lim_{hrightarrow 0}frac{sin(ha)}{h}

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