Please solve this
Q. $\underset{x\to \infty }{\lim } \frac{\sqrt{1+x^4}-\left(1+x^2\right)}{x^2}=$
(a) - 1
(b) 0
(c) 2
(d)

Dear Student,


$$\begin{gathered} \underset{x\to \infty }{\lim } \frac{\sqrt{1 + x^4} - \left(1 + x^2\right)}{x^2} \\ = \underset{x\to \infty }{\lim } \frac{\sqrt{1 + x^4}}{x^2} - \left(\frac{1 + x^2}{x^2}\right) \\ = \underset{x\to \infty }{\lim } \sqrt{\frac{1 + x^4}{x^4}} - \frac{1}{x^2} - 1 \\ = \underset{x\to \infty }{\lim } \sqrt{\frac{1}{x^4} + 1} - \frac{1}{x^2} - 1 \\ = \sqrt{0 + 1} - 0 - 1 \left[\because As x \to \infty , \frac{1}{x} \to 0 \Rightarrow \frac{1}{x^2} \to 0 and \frac{1}{x^4} \to 0\right] \\ = \sqrt{1} - 1 \\ = 1 - 1 \\ = 0 \end{gathered}$$

Hope this information will clear your doubts about topic.

​If you have any more doubts, please ask here on the forum and our experts will try and help you out as soon as possible.

​Regards