$$\lim_{x\rightarrow -\infty} (\sqrt{x^2+1 } -\sqrt{x^2-4x}) $$
The $+1$ and $-4x$ are not attached to the exponents.
Evaluate or explain why the limit does not exists.
2026-03-28 14:37:48.1774708668
Evaluate the limit without using the L'Hospital's Rule, if it is applicable.
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I'll get you started but you can finish where I leave off. Whenever you see square roots with limits you should immediately think to reverse rationalize:
$$\sqrt{x^2+1}-\sqrt{x^2-4x} = \frac{(\sqrt{x^2+1}-\sqrt{x^2-4x})(\sqrt{x^2+1}+\sqrt{x^2-4x})}{\sqrt{x^2+1}+\sqrt{x^2-4x}}.$$
The numerator becomes $(x^2+1)-(x^2-4x) = 4x+1$, giving
$$\lim_{x\to-\infty}\sqrt{x^2+1}-\sqrt{x^2-4x} = \lim_{x\to-\infty}\frac{4x+1}{\sqrt{x^2+1}+\sqrt{x^2-4x}}.$$
If you factor out $x^2$ from the square roots, you can easily evaluate this limit.