MCQ
$\lim _{x \rightarrow \infty}\left(\sqrt{x^2+8 x+3}-\sqrt{x^2+4 x+3}\right)=$
- A$0$
- B$\infty$
- ✓2
- D$\frac{1}{2}$
✓
Answer
Correct option: C.
2
(C)
On rationalising, we get
$\lim _{x \rightarrow \infty}\left(\sqrt{x^2+8 x+3}-\sqrt{x^2+4 x+3}\right)$
$=\lim _{x \rightarrow \infty} \frac{4 x}{\sqrt{x^2+8 x+3}+\sqrt{x^2+4 x+3}}$
$=\lim _{x \rightarrow \infty} \frac{4}{\left(\sqrt{1+\frac{8}{x}+\frac{3}{x^2}}+\sqrt{1+\frac{4}{x}+\frac{3}{x^2}}\right)}=2$
On rationalising, we get
$\lim _{x \rightarrow \infty}\left(\sqrt{x^2+8 x+3}-\sqrt{x^2+4 x+3}\right)$
$=\lim _{x \rightarrow \infty} \frac{4 x}{\sqrt{x^2+8 x+3}+\sqrt{x^2+4 x+3}}$
$=\lim _{x \rightarrow \infty} \frac{4}{\left(\sqrt{1+\frac{8}{x}+\frac{3}{x^2}}+\sqrt{1+\frac{4}{x}+\frac{3}{x^2}}\right)}=2$
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