Question

$\underset{x\to 0}{Lt}\frac{\sqrt{3+x^5}-\sqrt{3-x^5}}{\sin x}=$

• A. $0$
• B. $1$
• C. $1/\sqrt{3}$
• D. $1/\sqrt{5}$

Answer 1

Answer: (A)

$0$

Now,

$\underset{x\to 0}{Lt}\frac{\sqrt{3+x^5}-\sqrt{3-x^5}}{\sin x}$

$=\underset{x\to 0}{Lt}\frac{(\sqrt{3+x^5}{)}^2-(\sqrt{3-x^5}{)}^2}{\left(\sin x\right)(\sqrt{3+x^5}+\sqrt{3-x^5)}}$

$=\underset{x\to 0}{Lt}\frac{2x^5}{\left(\sin x\right)(\sqrt{3+x^5}+\sqrt{3-x^5)}}$

$=2\times \underset{x\to 0}{Lt}\frac{x^4}{(\sqrt{3+x^5}+\sqrt{3-x^5)}}\frac{1}{\underset{x\to 0}{lim}\frac{\sin x}{x}}$

$=2\times \frac{0}{2\sqrt{3}}.1$

$=0$.