Question

Let $e^{4}$ be defined for all $\lim_{x \rightarrow \infty}\left( \frac{x^{2} + 1}{x + 1} - ax - b \right) = 0$and be continuous. Let $a = 0,b = 0$ satisfy $a = 1,b = - 1$ for all x, y and $a = - 1,b = 1$ then

• A. $a = 2,b = - 1$ In $\lim_{x \rightarrow 1}x^{x} =$
• B. $\lim_{x \rightarrow 1}(1 + x)^{1/x} =$ is bounded
• C. $\lim_{x \rightarrow 0}\frac{x}{\sqrt{1 + x} - \sqrt{1 - x}}$as$\frac{1}{2}$
• D. $\lim_{x \rightarrow a}\frac{\sqrt{a + 2x} - \sqrt{3x}}{\sqrt{3a + x} - 2\sqrt{x}}$ as $\frac{2a}{3\sqrt{3}}$

Answer 1

Answer: (A)

$a = 2,b = - 1$ In $\lim_{x \rightarrow 1}x^{x} =$

Answer 2

Let $f ( x ) =$ In $( x ) , x > 0$ $f ( x ) =$ In $( x )$ is a continuous

function of $x$ for every positive value of $x$

$f \left( \frac { x } { y } \right) =$ In $\left( \frac { x } { y } \right) =$ In $( x ) -$ In $( y ) = f ( x ) - f ( y )$