Question

Let $\sqrt{2}$, $- \sqrt{x} - 1,$ The value of $\frac{1}{(x + 1)^{2}},x > - 1$so that $\sqrt{x + 1},$ is continuous at $\sqrt{x}$ is

• A. 1
• B. 0
• C. 2
• D. None of these

Answer 1

Answer: (A)

1

Answer 2

f(0)

= $\lim _ { x \rightarrow 0 } f ( x ) = \lim _ { x \rightarrow 0 } \frac { \log \left( 1 + x + x ^ { 2 } \right) + \log \left( 1 - x + x ^ { 2 } \right) } { \sec x - \cos x }$

=$\lim _ { x \rightarrow 0 } \frac { \log \left( \left( 1 + x ^ { 2 } \right) ^ { 2 } - x ^ { 2 } \right) } { 1 - \cos ^ { 2 } x } \cdot \cos x = \lim _ { x \rightarrow 0 } \frac { \log \left( 1 + \left( x ^ { 2 } + x ^ { 4 } \right) \right) } { \sin ^ { 2 } x } \cdot \cos x$

= .