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

1. ## 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

2. ## 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. . .

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

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

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

Divide numerator and denominator by $x\!:$

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

. . $=\;\lim_{x\to\infty}\dfrac{10+\dfrac{1}{x}}{\sqrt{ 1 + \dfrac{10}{x} + \dfrac{1}{x^2}} + 1} \;=\; \dfrac{10+0}{\sqrt{1+0+0} + 1} \;=\;\frac{10}{2} \;=\;5$

3. ## Re: Evaluate the limits - half the work already done, thanks

Originally Posted by Soroban
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?

4. ## Re: Evaluate the limits - half the work already done, thanks

I'm still waiting... thanks

5. ## Re: Evaluate the limits - half the work already done, thanks

Originally Posted by dokrbb
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 $\infty$.

6. ## Re: Evaluate the limits - half the work already done, thanks

Originally Posted by Plato
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

7. ## Re: Evaluate the limits - half the work already done, thanks

Originally Posted by dokrbb
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 $-\infty$.