$$ \lim_{x \to -\infty} \frac{4x}{\sqrt{16x^2 + 1}}$$ The result should be -1 but how
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1What have you tried so far ? – Digitallis Oct 03 '23 at 08:19
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Note that $u^2+1=u^2(1+u^{-2})$. – Gerry Myerson Oct 03 '23 at 08:27
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Welcome to the site! There are many similar problems to yours. For example check this. There are different ways to search, you can use search bar on here or you can use Approach Zero. – Etemon Oct 03 '23 at 08:33
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1I’m voting to close this question because OP is not engaging. – Gerry Myerson Oct 04 '23 at 21:40
1 Answers
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converting the limit from tending to +$\infty$ instead of -$\infty$ makes it easier to work with, so that you don't commit errors and get 1 as the answer.
convert to:
$\lim_{x\to \infty}\frac{-4x}{\sqrt{16x^2+1}}$
now it's easy. just divide both the numerator and denominator by 4x so we divide under the square root and get
$\lim_{x\to \infty}\frac{-1}{\sqrt{1 +{\frac{1}{16x^2}}}}$
= -1
rinn
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