$$\lim\limits_{n\to \infty}\left(\sqrt{x^2+4x\ +1}-x\right)$$
The answer is $2$ but I don't know how to evaluate it
Asked
Active
Viewed 80 times
-1
-
2But theres nothing on rhs no equality or is it a limits question – Archis Welankar Mar 06 '16 at 18:00
-
1I don't understand. Do you want a limit as $x\to\text{something}$ of that function? If so, what's "$\text{something}$"? Also, share with us your attempts. – Mar 06 '16 at 18:00
-
1I assume he means $\lim\limits_{x \to \infty} $ of that given equation, it would fit the answer. – Imago Mar 06 '16 at 18:05
-
3Duplicate: $\lim\limits_{x\rightarrow \infty}\sqrt[n]{x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}}-x$. – Workaholic Mar 06 '16 at 18:13
-
Hint: $\sqrt{x^2+4x+1}=x\sqrt{1+4/x+1/x^2}\approx x(1+2/x+1/2x^2)$. – Mar 06 '16 at 18:18
-
@YvesDaoust there is a minus $x$ – 3SAT Mar 06 '16 at 18:20
-
Multiply by $A/A$ where $A=\sqrt {x^2+4 x+1}+x$. – DanielWainfleet Mar 06 '16 at 18:43
-
Completing the square is the easiest way to see it, I think. – Vik78 Mar 06 '16 at 18:59
-
@Nehorai: I know, I didn't give the complete solution. – Mar 06 '16 at 19:31
-
@Lucas andre don't forget to accept one of the answers clicking the 'V' in the left side of the answer, and also to vote – 3SAT Mar 08 '16 at 19:30
2 Answers
1
$$\lim _{x \rightarrow \infty} \left(\left(\sqrt{x^2+4x\ +1}\right)-x\right)=\lim _{x \rightarrow \infty}\frac{\left(x^2+4x +1\right)-x^2}{\left(\sqrt{x^2+4x\ +1}\right)+x}=$$ $$\lim _{x \rightarrow \infty}=\frac{4x +1}{\sqrt{x^2+4x\ +1}+x}=\frac 42=2$$
Roman83
- 17,884
- 3
- 26
- 70
0
$$=\lim _{x \rightarrow \infty} \frac{\left(\sqrt{x^2+4x\ +1}-x\right)}{1}$$
Multiply nominator and denominator by $\sqrt{x^2+4x\ +1}+x$
$$\require{cancel}=\lim _{x \rightarrow \infty}\frac{\left(\cancel{x^2}+4x +1\right)\cancel{-x^2}}{\left(\sqrt{x^2+4x\ +1}\right)+x}$$ $$=\lim _{x \rightarrow \infty}\frac{4x +1}{\sqrt{x^2+4x\ +1}+x}$$
Divide nominator and denominator by $x$
$$=\lim _{x \rightarrow \infty}\frac{4 +\overbrace{1/x}^{\to 0}}{\sqrt{1+\underbrace{4/x}_{\to 0} +\underbrace{1/x^2}_{\to0}}+1}=\frac {4} {1+1}=\color{red}2$$
3SAT
- 7,512