Question 1200146
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When x < 0, {{{sqrt(x^2) = -x}}}


When x > 0, {{{sqrt(x^2) = x}}}


For very large positive or negative values of x, {{{sqrt(x^2-1)}}} approaches {{{sqrt(x^2)}}}. 


This can be seen in the graph shown below
<a href = "https://www.desmos.com/calculator/1llyy35ent">https://www.desmos.com/calculator/1llyy35ent</a>
Desmos is a free graphing app.


The -1 at the end doesn't alter the radicand too much when x^2 is so very large.
Examples:
If x = 5, then x^2 = 25 and x^2-1 = 24
If x = 50, then x^2 = 2500 and x^2-1 = 2499
If x = 500, then x^2 = 250,000 and x^2-1 = 249,999
If x = 5000, then x^2 = 25,000,000 and x^2-1 = 24,999,999


This means {{{sqrt(x^2-1)}}} approaches {{{-x}}} for large negative values of x.
Also, {{{sqrt(x^2-1)}}} approaches {{{x}}} for large positive values of x.



For very large negative values of x, {{{x/sqrt(x^2-1)}}} approaches {{{x/(-x) = -1}}}


For very large positive values of x, {{{x/sqrt(x^2-1)}}} approaches {{{x/x = 1}}}


Therefore,
*[tex \Large \displaystyle \lim_{\text{x}\to -\infty} \left( \frac{\text{x}}{\sqrt{\text{x}^2-1}}\right) = -1]
and
*[tex \Large \displaystyle \lim_{\text{x}\to \infty} \left(\frac{\text{x}}{\sqrt{\text{x}^2-1}}\right) = 1]


Verification using WolframAlpha
<a href = "https://www.wolframalpha.com/input?i=x%2Fsqrt%28x%5E2-1%29+asymptotes">https://www.wolframalpha.com/input?i=x%2Fsqrt%28x%5E2-1%29+asymptotes</a>


Verification using Desmos
<a href = "https://www.desmos.com/calculator/hfq1eravms">https://www.desmos.com/calculator/hfq1eravms</a>
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