I am interested in evaluation $\lim_{x\to ∞} \left(\sqrt{x^2+4x+1}-x \right)$.
I tried to multiply it by ($\sqrt{x^2+4x+1}+x$) but didn't seem like it did worked.
Asked
Active
Viewed 68 times
0
-
You can write \infty for the $\infty$ symbol. :) – layman Nov 13 '16 at 19:12
-
2Try multiplying by $\frac{\sqrt{x^2+4x+1}+x}{\sqrt{x^2+4x+1}+x}$ instead – Sil Nov 13 '16 at 19:12
-
5Famous duplicate: $\displaystyle\lim_{x\to\infty}\sqrt[n]{x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}}-x$. See also this one. – Workaholic Nov 13 '16 at 19:32
2 Answers
3
$$\lim_{x \to \infty} \sqrt{x^2+4x+1}-x = \lim_{x \to \infty} \frac{x^2+4x+1-x^2}{\sqrt{x^2+4x+1}+x}= \lim_{x \to \infty} \frac{4x+1}{|x|\sqrt{1+\frac{4}{x}+\frac{1}{x^2}}+x}\\= \lim_{x \to \infty} \frac{4+\frac{1}{x}}{\sqrt{1+\frac{4}{x}+\frac{1}{x^2}}+1}$$
N. S.
- 132,525
1
I thought it might be instructive to present a way forward that does not rationalize the term of interest. Rather, we rely on the Generalized Binomial Theorem. To that end, we proceed.
Applying the Generalized Binomial Theorem to the square root reveals
$$\begin{align} \lim_{x\to \infty}\left(\sqrt{x^2+4x+1}-x\right)&=\lim_{x\to \infty}\left(x\left(1+\frac{4x+1}{x^2}\right)^{1/2}-x\right)\\\\ &=\lim_{x\to \infty}\left(\frac{4x+1}{2x}+O\left(\frac1{x}\right)\right)\\\\ &=2 \end{align}$$
Mark Viola
- 179,405