Question

pls ans it fast
Class 11 ​

Answer 1

Step-by-step explanation:

Lim $\: \frac{ \sqrt{x} }{ \sqrt{x + \sqrt{x + \sqrt{x} } } }$

Lim $\frac{1}{ \frac{ \sqrt{x + \sqrt{x + \sqrt{x} } } }{ \sqrt{x} } }$

Lim $\frac{1}{ \sqrt{ \frac{x + \sqrt{x + \sqrt{x} } }{x} } }$

Lim $\frac{1}{ \sqrt{ \frac{x}{x} + \frac{ \sqrt{x + \sqrt{x} } }{x} } }$

Lim $\frac{1}{ \sqrt{1 + \frac{ \sqrt{x + \sqrt{x} } }{ \sqrt{ {x}^{2} } } } }$

Lim $\frac{1}{ \sqrt{1 + \sqrt{ \frac{x + \sqrt{x} }{ {x}^{2} } } } }$

Lim $\frac{1}{ \sqrt{1 + \sqrt{ \frac{x}{ {x}^{2} } + \frac{ \sqrt{x} }{ {x}^{2} } } } }$

Lim $\frac{1}{ \sqrt{1 + \sqrt{ \frac{1}{x} + \frac{1}{x \sqrt{x} } } } }$

Lim $\frac{1}{ \sqrt{1 + \sqrt{ \frac{1}{x} + \frac{1}{ {x}^{3/2 } } } } }$

Putting value

As x tends to infinity, $\frac{1}{{x}^{n}}$ tends to zero

$=\frac{1}{ \sqrt{1 + \sqrt{0 + 0} } }$

$=\frac{1}{ \sqrt{1} }$

$=1$

Hope I'm correct