This question has four choices (A), (B), (C) and (D) out of which ONE or MORE
are correct.

Let $f (x) = x + \sqrt{x^{2} + 2 x} a n d g (x) = \sqrt{x^{2} + 2 x} - x$ , then

l i m x \to \infty g (x) = 1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

l i m x \to - \infty f (x) = 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

l i m x \to - \infty f (x) = - 1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

l i m x \to \infty g (x) = - 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct options are
A $l i m x \to \infty g (x) = 1$
C $l i m x \to - \infty f (x) = - 1$
$l i m x \to \infty g (x) = l i m x \to \infty \frac{(\begin{matrix} \sqrt{x^{2} + 2 x} - x \end{matrix}) (\begin{matrix} \sqrt{x^{2} + 2 x} + x \end{matrix})}{\sqrt{x^{2} + 2 x} + x} = 1$

$l i m x \to - \infty f (x) = l i m x \to - \infty \frac{(\begin{matrix} \sqrt{x^{2} + 2 x} - x \end{matrix}) (\begin{matrix} \sqrt{x^{2} + 2 x} + x \end{matrix})}{\sqrt{x^{2} + 2 x} - x}$

$= l i m x \to - \infty \frac{2 x}{- x \sqrt{1 + \frac{2}{x}} - x} = l i m x \to - \infty \frac{2}{- \sqrt{1 + \frac{2}{x}} - 1} = - 1$

Suggest Corrections

Similar questions

\sqrt{25 - x^{2}}

lim x \to 1 \frac{G (x) - G (1)}{x - 1}

f (x) = - \sqrt{25 - x^{2}}

lim x \to 1 \frac{f (x) - f (1)}{x - 1}

f (x) = \frac{a x + b}{x + 1}, lim x \to \infty f (x) = 1

lim x \to 0 f (x) = 2

f (- 2)

f : [- \frac{1}{2}, \infty) \to R

f (x) = \sqrt{1 + 2 x}, x \in [- \frac{1}{2}, \infty)

lim x \to ⎛ ⎝ \frac{1}{2} ⎞ ⎠ f (x) .

lim x \to \frac{- 1}{2} f (x)

l i m x \to a f (x) g (x)

l i m x \to a f (x)

l i m x \to a g (x)

Join BYJU'S Learning Program