Computing Limits at Infinity Part III

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Hckyplayer8

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One last one for tonight.

Compute the lim as x--> neg infinity of [sqrt(2x2+1)] / (x+4)

The x2 under the radical is negating the power, correct? So that means the variables are actually equivalent to one another sans one is being multiplied by the constant 2.

So after dividing the given by x I have sqrt[2 + (1/x)] / [1 + (4/x] thus far.

Is that correct?
 
Hckyplayer8 said:
One last one for tonight.

Compute the lim as x--> neg infinity of [sqrt(2x2+1)] / (x+4)

The x2 under the radical is negating the power, correct? So that means the variables are actually equivalent to one another sans one is being multiplied by the constant 2.

So after dividing the given by x I have sqrt[2 + (1/x2)] / [1 + (4/x] thus far. Note exponent of 2 added in.

Is that correct?
Click to expand...
see correction above
You are dividing numerator and denominator by x or sqrt(x2).
 
Don't forget that since [MATH]x[/MATH] is approaching [MATH]-\infty[/MATH], not [MATH]+\infty[/MATH], you have to take [MATH]x[/MATH] as negative. In dividing the numerator by [MATH]\sqrt{x^2}[/MATH], you are dividing the numerator by [MATH]|x|[/MATH] but the denominator by [MATH]x[/MATH]; these are not the same!
 
Could someone please walk me through this?

I tried backtracking to figure out why the variable needed the exponent but it was to no avail. Why is the numerator and denominator divided differently. I thought you divide the equation by the highest power in the denominator?
 
Note that because x is approaching negative infinity, x is negative (see Dr Ps post above).

\(\displaystyle \lim_{x\to -\infty} \frac{\sqrt{2x^2+1}}{x+4}\)

=\(\displaystyle \lim_{x\to -\infty} \frac{\sqrt{2x^2+1}/|x|}{(x+4)/ |x|}\)

Now when you take the |x| into the square root sign it becomes x^2.
 
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Harry_the_cat said:
Note that because x is approaching negative infinity, x is negative (see Dr Ps post above).

\(\displaystyle \lim_{x\to -\infty} \frac{\sqrt{2x^2+1}}{x+4}\)

=\(\displaystyle \lim_{x\to -\infty} \frac{\sqrt{2x^2+1}/|x|}{(x+4)/ |x|}\)

Now when you take the |x| into the square root sign it becomes x^2.
Click to expand...

Thank you for replying. I apologize but I still don't understand. I must have data dumped something. Why is the variable being transitioned to an absolute value?

I understand that the variable is going towards negative infinity.
 
I (almost) never compute limits as x approaches neg infinity. I simply replace x with -x and then take the limit as x goes to positive infinity.
 
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