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.

Question

anonymous · Answer

$$\lim_{x \rightarrow -\infty} \frac{ \sqrt{x ^2 + 8} -3 }{ 1- x }$$

myininaya · Answer

Multiply top and bottom by $$1/\sqrt{x^2}$$

myininaya · Answer

Now since x-> inf and not -inf
then |x|=x not -x
So we have
$$\lim_{x \rightarrow \infty}\frac{\sqrt{x^2+8}-3}{1-x} \frac{\frac{1}{\sqrt{x^2}}}{\frac{1}{x}}$$

anonymous · Answer

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 :-)

myininaya · Answer

$$\lim_{x \rightarrow \infty}\frac{\sqrt{\frac{x^2+8}{x^2}}-\frac{3}{x}}{\frac{1}{x}-1}$$

myininaya · Answer

the answer shouldn't be 1
...

anonymous · Answer

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.

anonymous · Answer

You forgot something. The x before those expressions.

anonymous · Answer

Now I am getting a -1, as you are implying. Thats weird. I need to recheck.

myininaya · Answer

What do you mean the x before the expressions?

myininaya · Answer

right -1 is what i got

anonymous · Answer

You have limit goes to infinity. It should go to -infinity. Im now checking whether it has some impact on the solution...

myininaya · Answer

So the question was suppose to be
 $$\lim_{x \rightarrow - \infty}\frac{\sqrt{x^2+8}-3}{1-x}$$

anonymous · Answer

Yes. I typed it right the first time.

myininaya · Answer

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.

myininaya · Answer

$$|x|=-x \text{ if } x<0$$

anonymous · Answer

But I still get a -1. I checked with Wolfram it gets +1.

anonymous · Answer

Absolute value, Im following. Got the same.

myininaya · Answer

$$\lim_{x \rightarrow - \infty}\frac{\sqrt{x^2+8}-3}{1-x} \cdot \frac{\frac{1}{\sqrt{x^2}}}{\frac{1}{\sqrt{x^2}}}$$
$$\lim_{x \rightarrow -\infty}\frac{\sqrt{\frac{x^2+8}{x^2}}-\frac{3}{-x}}{\frac{1}{-x}-\frac{x}{-x}}$$

myininaya · Answer

Yep the limit is 1 if x->-inf

anonymous · Answer

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.

myininaya · Answer

lol. Thanks.