I was watching this lecture on limits and the lecturer did the following algebraic manipulation:

$$ \begin{align*} &= \lim_{x \to \infty} \frac{ \sqrt{x^2-4x+1} + x }{1-4x} . \frac{ \frac{1}{x} }{ \frac{1}{x} } \\ &= \lim_{x \to \infty} \frac{ \sqrt{1 - \frac{4}{x} + \frac{1}{x^2} } + 1 }{ \frac{1}{x} - 4 } \end{align*} $$

The part I didn't understand is how $\frac{1}{x}$ multiplied with $\sqrt{x^2-4x+1}$.


Well, we know that as long as $x$ is positive, then $\sqrt{\frac{1}{x^2}}= \frac{1}{x}$
We also know that $\sqrt{a} \times \sqrt{b} = \sqrt{ab}$
Thus, $\frac{1}{x} \times \sqrt{x^2-4x+1} = \sqrt{\frac{1}{x^2}} \times \sqrt{x^2-4x+1} = \sqrt{\frac{1}{x^2} \times (x^2-4x+1)} = \sqrt{1-\frac{4}{x}+\frac{1}{x^2}}$
I believe you have a typo and accidentally wrote $-\frac{1}{x^2}$.

  • $\begingroup$ Check the first term in your final expression. $\endgroup$ – N. F. Taussig Aug 12 '15 at 9:46
  • $\begingroup$ Thank you! Sorry, I'm not used to LaTeX. $\endgroup$ – Elie Louis Aug 12 '15 at 9:48
  • $\begingroup$ Thank you for the explanation.. and pointing out the typo. $\endgroup$ – Daniyal Aug 12 '15 at 9:53

Hint: $$\frac{1}{x} = \sqrt{\frac{1}{x^2}}$$ and for any $a,b \in \mathbb{R}^{\geq 0}$ $$\sqrt{a}\sqrt{b} = \sqrt{ab}$$

  • $\begingroup$ Thanks.. didn't thought about hint before. $\endgroup$ – Daniyal Aug 12 '15 at 9:54

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