One limit keeps slipping through my fingers. lim_{x rightarrow -infty} \frac{ sqrt{x ^2 + 8} -3 }{ 1- x } . The answer =1, but I keep getting =2. I will be posting my steps, please tell me when I'm wrong.

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One limit keeps slipping through my fingers. lim_{x rightarrow -infty} \frac{ sqrt{x ^2 + 8} -3 }{ 1- x } . The answer =1, but I keep getting =2. I will be posting my steps, please tell me when I'm wrong.

Calculus1
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\[\lim_{x \rightarrow -\infty} \frac{ \sqrt{x ^2 + 8} -3 }{ 1- x }\]
Multiply top and bottom by \[1/\sqrt{x^2}\]
Now since x-> inf and not -inf then |x|=x not -x So we have \[\lim_{x \rightarrow \infty}\frac{\sqrt{x^2+8}-3}{1-x} \frac{\frac{1}{\sqrt{x^2}}}{\frac{1}{x}}\]

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Oh, Ive got it... Im sorry everyone. Now I see, I was making one very silly mistake. I didnt have the top part - I did not multiply the -3, which got me the wrong direction.. Thanks for the answer @myiniaya . My bad :-)
\[\lim_{x \rightarrow \infty}\frac{\sqrt{\frac{x^2+8}{x^2}}-\frac{3}{x}}{\frac{1}{x}-1}\]
the answer shouldn't be 1 ...
Yes. I had the same as you in your last post, but I forgot the and put there only -3 instead. Which was really silly.
You forgot something. The x before those expressions.
Now I am getting a -1, as you are implying. Thats weird. I need to recheck.
What do you mean the x before the expressions?
right -1 is what i got
You have limit goes to infinity. It should go to -infinity. Im now checking whether it has some impact on the solution...
So the question was suppose to be \[\lim_{x \rightarrow - \infty}\frac{\sqrt{x^2+8}-3}{1-x}\]
Yes. I typed it right the first time.
I completely didn't see the negative infinity when I read it the first time as you can see in that one comment I made above.
\[|x|=-x \text{ if } x<0\]
But I still get a -1. I checked with Wolfram it gets +1.
Absolute value, Im following. Got the same.
\[\lim_{x \rightarrow - \infty}\frac{\sqrt{x^2+8}-3}{1-x} \cdot \frac{\frac{1}{\sqrt{x^2}}}{\frac{1}{\sqrt{x^2}}}\] \[\lim_{x \rightarrow -\infty}\frac{\sqrt{\frac{x^2+8}{x^2}}-\frac{3}{-x}}{\frac{1}{-x}-\frac{x}{-x}}\]
Yep the limit is 1 if x->-inf
Thank you very much. I completely lost the one '-' sign. You cleared everything out. And btw. I dont understand how come you are so fast with putting such expressions through the keyboard on the screen. Youve earned a medal from me. Thank you.
lol. Thanks.

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