$$ \left. \begin{array} { l } { f = + \infty \quad f ( x ) = \frac { \sqrt { x ^ { 2 } + 1 } - x } { \sqrt { x ^ { 2 } - 1 } - \sqrt { x ^ { 2 } - 2 } } } \ { \lim _ { x ightarrow + \infty } f ( x ) = } \end{array} ight. $$

Question

$$ \left. \begin{array} { l } { f = + \infty \quad f ( x ) = \frac { \sqrt { x ^ { 2 } + 1 } - x } { \sqrt { x ^ { 2 } - 1 } - \sqrt { x ^ { 2 } - 2 } } } \ { \lim _ { x ightarrow + \infty } f ( x ) = } \end{array} ight. $$

UpStudy AI · Accepted Answer

We start with
$$
f(x)=\frac{\sqrt{x^{2}+1}-x}{\sqrt{x^{2}-1}-\sqrt{x^{2}-2}}.
$$

**Step 1. Simplify the numerator**

Rationalize the numerator by multiplying by the conjugate:
$$
\sqrt{x^{2}+1}-x=\frac{(\sqrt{x^{2}+1}-x)(\sqrt{x^{2}+1}+x)}{\sqrt{x^{2}+1}+x}=\frac{(x^{2}+1)-x^{2}}{\sqrt{x^{2}+1}+x}=\frac{1}{\sqrt{x^{2}+1}+x}.
$$
For large $$x$$, we have
$$
\sqrt{x^{2}+1}\approx x,
$$
so
$$
\sqrt{x^{2}+1}+x\approx 2x \quad \text{and} \quad \sqrt{x^{2}+1}-x\approx\frac{1}{2x}.
$$

**Step 2. Simplify the denominator**

Similarly, rationalize the denominator by multiplying by its conjugate:
$$
\sqrt{x^{2}-1}-\sqrt{x^{2}-2}=\frac{(\sqrt{x^{2}-1}-\sqrt{x^{2}-2})(\sqrt{x^{2}-1}+\sqrt{x^{2}-2})}{\sqrt{x^{2}-1}+\sqrt{x^{2}-2}}.
$$
Notice that
$$
(\sqrt{x^{2}-1}-\sqrt{x^{2}-2})(\sqrt{x^{2}-1}+\sqrt{x^{2}-2})=(x^{2}-1)-(x^{2}-2)=1.
$$
Hence,
$$
\sqrt{x^{2}-1}-\sqrt{x^{2}-2}=\frac{1}{\sqrt{x^{2}-1}+\sqrt{x^{2}-2}}.
$$
For large $$x$$, we have
$$
\sqrt{x^{2}-1}\approx x \quad \text{and} \quad \sqrt{x^{2}-2}\approx x,
$$
so
$$
\sqrt{x^{2}-1}+\sqrt{x^{2}-2}\approx 2x \quad \text{and} \quad \sqrt{x^{2}-1}-\sqrt{x^{2}-2}\approx\frac{1}{2x}.
$$

**Step 3. Compute the limit**

Substitute the asymptotic forms into $$f(x)$$:
$$
f(x)\approx\frac{\frac{1}{2x}}{\frac{1}{2x}}.
$$
Thus,
$$
\lim_{x\rightarrow+\infty}f(x)=1.
$$
The limit of $$ f(x) $$ as $$ x $$ approaches infinity is 1.

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