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Math Help - Evaluate the limits - half the work already done, thanks

  1. #1
    Member dokrbb's Avatar
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    Evaluate the limits - half the work already done, thanks

    Can you please show me how all the algebra works for evaluating the limits of this function as x-> +/_ infinity for this function

    lim f(x),x-> +/_ inf sqrt(x^2 + 10x +1) x

    That's what I tried, and I figured out the first asymptote:

    lim f(x),x-> +/_ inf sqrt(x^2 + 10x +1) x=

    lim f(x),x-> +/_ inf sqrt(x^2 + 10x +1) x*([sqrt(x^2 + 10x +1) +x]/[sqrt(x^2 + 10x +1) +x])=

    lim f(x),x-> +/_ inf [(x^2 + 10x +1) x^2]/sqrt(x^2 + 10x +1) + x =

    a)
    divide by x both the numerator and the denominator:

    lim f(x),x-> + inf [(x + 10 + 0) x]/sqrt(1 + 0 +0) + 1 = 10/2 = 5

    this would be the horizontal asymtote to which f(x) approaches when x becomes large positive,

    b) but what happens when x becomes large negative? I see from the graph that f(x) tends to positive infinitive, but I can't fugure it out algebraically.

    lim f(x),x-> - inf [(x^2 + 10x +1) x^2]/sqrt(x^2 + 10x +1) + x = ...

    Thanks
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    Re: Evaluate the limits - half the work already done, thanks

    Hello, dokrbb!

    \lim_{x\to \infty}\left[\sqrt{x^2 + 10x +1} x \right]

    That's what I tried. . .

    \lim_{x\to\infty}\left[\sqrt{x^2 + 10x +1} x\right]

    . . =\;\lim_{x\to\infty}\frac{\sqrt{x^2+10x+1} - x}{1}\cdot \frac{\sqrt{x^2+10x+1} + x}{\sqrt{x^2+10x+1} + x }

    . . =\;\lim_{x\to\infty}\frac{(x^2+10x+1) - x^2}{\sqrt{x^2+10x+1} + x} \;=\;\lim_{x\to\infty}\frac{10x+1}{\sqrt{x^2+10x+1  } + x}

    I don't follow your steps after this.

    Divide numerator and denominator by x\!:

    \lim_{x\to\infty}\dfrac{\dfrac{10x+1}{x}}{\dfrac{ \sqrt{x^2+10x+1} + x}{x}} \;=\;\lim_{x\to\infty}\dfrac{\dfrac{10x}{x} + \dfrac{1}{x}}{\sqrt{\dfrac{x^2+10x+1}{x^2}} + \dfrac{x}{x}}

    . . =\;\lim_{x\to\infty}\dfrac{10+\dfrac{1}{x}}{\sqrt{  1 + \dfrac{10}{x} + \dfrac{1}{x^2}} + 1} \;=\; \dfrac{10+0}{\sqrt{1+0+0} + 1} \;=\;\frac{10}{2} \;=\;5
    Thanks from dokrbb
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    Re: Evaluate the limits - half the work already done, thanks

    Quote Originally Posted by Soroban View Post
    Hello, dokrbb!


    Divide numerator and denominator by x\!:

    \lim_{x\to\infty}\dfrac{\dfrac{10x+1}{x}}{\dfrac{ \sqrt{x^2+10x+1} + x}{x}} \;=\;\lim_{x\to\infty}\dfrac{\dfrac{10x}{x} + \dfrac{1}{x}}{\sqrt{\dfrac{x^2+10x+1}{x^2}} + \dfrac{x}{x}}

    . . =\;\lim_{x\to\infty}\dfrac{10+\dfrac{1}{x}}{\sqrt{  1 + \dfrac{10}{x} + \dfrac{1}{x^2}} + 1} \;=\; \dfrac{10+0}{\sqrt{1+0+0} + 1} \;=\;\frac{10}{2} \;=\;5
    Isn't this the "Divide by the highest power of x", what's wrong with it?
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    Re: Evaluate the limits - half the work already done, thanks

    I'm still waiting... thanks
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    Re: Evaluate the limits - half the work already done, thanks

    Quote Originally Posted by dokrbb View Post
    I'm still waiting...
    For what are you still waiting?

    Soroban gave you the complete solution for the case x\to\infty, even though you had correctly done it in the OP.

    In the case x\to -\infty, the expression \sqrt{x^2+10x+1}-x\ge 10 for x<-10 and it grows rapidly.
    So the limit is \infty.
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    Re: Evaluate the limits - half the work already done, thanks

    Quote Originally Posted by Plato View Post
    In the case x\to -\infty, the expression \sqrt{x^2+10x+1}-x\ge 10 for x<-10 and it grows rapidly.
    So the limit is \infty.
    That's exactly what I don't get, can you please show me this by steps,

    I would be grateful, really
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    Re: Evaluate the limits - half the work already done, thanks

    Quote Originally Posted by dokrbb View Post
    That's exactly what I don't get, can you please show me this by steps,
    I really do not know what more to say.
    Look at this graph

    For any x<-10 then -x>10 and \sqrt{x^2+10x+1}>0. So the sum increases to -\infty.
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