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$$\lim_{x \to \infty} (x+2)-\sqrt{x^2+6x-1}$$

I've tried multiplying by conjugate and dividing by $x$ but still get $0$ in the denominator.

Mithlesh Upadhyay
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Blake
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    You need to check the formatting on this question. I can't see a denominator at all. It would also help if you wrote out the work you have done, so we can check where the mistake is. – user141592 Oct 10 '14 at 04:10
  • i multiplied by the conjugate and divided everything by x sorry thats where my denominator came from – Blake Oct 10 '14 at 04:12
  • wait i just found my mistake in the denominator a had a minus instead of plus giving me 1-1 in the denominator. so the answer would be -1 correct? – Blake Oct 10 '14 at 04:15
  • Typically the denominator is a sum of two otherwise-non-combinable terms. Could you state the orginal question? – abiessu Oct 10 '14 at 04:21
  • "I've tried multiplying by conjugate" Indeed this works nicely and yields the limit $-1$. – Did Oct 10 '14 at 05:54

2 Answers2

2

$(x+2)-\sqrt{x^2+6x-1}=\frac{-2x+5}{(x+2)+\sqrt{x^2+6x-1}}=\frac{-2+\frac{5}{x}}{(1+\frac{2}{x})+\sqrt{1+\frac{6}{x}-\frac{1}{x^2}}}$

As $x$ goes to $\infty$, we have limit $-1$.

John
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Setting $\dfrac1x=h,$

$$\lim_{x \to \infty} (x+2)-\sqrt{x^2+6x-1}$$

$$=\lim_{h\to0^+}\frac{1+2h-\sqrt{1+6h-h^2}}h$$

$$=\lim_{h\to0^+}\frac{(1+2h)^2-(1+6h-h^2)}{h(1+2h+\sqrt{1+6h-h^2})}$$

$$=\lim_{h\to0^+}\frac{h(5h-2)}{h(1+2h+\sqrt{1+6h-h^2})}$$

As $h\to0,h\ne0;$ safely cancel out $h$

Finally set $h=0$

lab bhattacharjee
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