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Mark_CZE

  • 3 years ago

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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  1. Mark_CZE
    • 3 years ago
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    \[\lim_{x \rightarrow -\infty} \frac{ \sqrt{x ^2 + 8} -3 }{ 1- x }\]

  2. myininaya
    • 3 years ago
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    Multiply top and bottom by \[1/\sqrt{x^2}\]

  3. myininaya
    • 3 years ago
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    Now since x-> inf and not -inf then |x|=x not -x So we have \[\lim_{x \rightarrow \infty}\frac{\sqrt{x^2+8}-3}{1-x} \frac{\frac{1}{\sqrt{x^2}}}{\frac{1}{x}}\]

  4. Mark_CZE
    • 3 years ago
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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 :-)

  5. myininaya
    • 3 years ago
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    \[\lim_{x \rightarrow \infty}\frac{\sqrt{\frac{x^2+8}{x^2}}-\frac{3}{x}}{\frac{1}{x}-1}\]

  6. myininaya
    • 3 years ago
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    the answer shouldn't be 1 ...

  7. Mark_CZE
    • 3 years ago
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    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.

  8. Mark_CZE
    • 3 years ago
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    You forgot something. The x before those expressions.

  9. Mark_CZE
    • 3 years ago
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    Now I am getting a -1, as you are implying. Thats weird. I need to recheck.

  10. myininaya
    • 3 years ago
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    What do you mean the x before the expressions?

  11. myininaya
    • 3 years ago
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    right -1 is what i got

  12. Mark_CZE
    • 3 years ago
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    You have limit goes to infinity. It should go to -infinity. Im now checking whether it has some impact on the solution...

  13. myininaya
    • 3 years ago
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    So the question was suppose to be \[\lim_{x \rightarrow - \infty}\frac{\sqrt{x^2+8}-3}{1-x}\]

  14. Mark_CZE
    • 3 years ago
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    Yes. I typed it right the first time.

  15. myininaya
    • 3 years ago
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    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.

  16. myininaya
    • 3 years ago
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    \[|x|=-x \text{ if } x<0\]

  17. Mark_CZE
    • 3 years ago
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    But I still get a -1. I checked with Wolfram it gets +1.

  18. Mark_CZE
    • 3 years ago
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    Absolute value, Im following. Got the same.

  19. myininaya
    • 3 years ago
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    \[\lim_{x \rightarrow - \infty}\frac{\sqrt{x^2+8}-3}{1-x} \cdot \frac{\frac{1}{\sqrt{x^2}}}{\frac{1}{\sqrt{x^2}}}\] \[\lim_{x \rightarrow -\infty}\frac{\sqrt{\frac{x^2+8}{x^2}}-\frac{3}{-x}}{\frac{1}{-x}-\frac{x}{-x}}\]

  20. myininaya
    • 3 years ago
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    Yep the limit is 1 if x->-inf

  21. Mark_CZE
    • 3 years ago
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    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.

  22. myininaya
    • 3 years ago
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    lol. Thanks.

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