Find a limit that is approaching infinity: (x*sqrt{x} + 4)/(2x^2 + 3x - 6)

[coooool222]

The answer is 1/2 but why the heck am i getting 0

work:

x/x^2 * (sqrt(x2/x4 + 4/x^2)/(2x^2)/x^2 + 3x/x^2 - 6/x^2

I plug in infinity and i get 0/2

 

[Dr.Peterson]

View attachment 37059

The answer is 1/2 but why the heck am i getting 0

work:

x/x^2 * (sqrt(x2/x4 + 4/x^2)/(2x^2)/x^2 + 3x/x^2 - 6/x^2

I plug in infinity and i get 0/2

Your work isn't very clear (an image of handwriting would be better), but I think what you mean by it is mostly correct.

I think your work is this: [math]\lim_{x\to+\infty}\frac{\frac{x}{x^2}\sqrt{\frac{x^2}{x^4}+\frac{4}{x^4}}}{\frac{2x^2}{x^2}+\frac{3x}{x^2}-\frac{6}{x^2}}[/math]
The main problem is that you divided the numerator by [imath]x^2[/imath] twice. Do you see that?

(I corrected one typo in addition to lots of missing parentheses and carets.)

 

[coooool222]

Your work isn't very clear (an image of handwriting would be better), but I think what you mean by it is mostly correct.

I think your work is this: [math]\lim_{x\to+\infty}\frac{\frac{x}{x^2}\sqrt{\frac{x^2}{x^4}+\frac{4}{x^4}}}{\frac{2x^2}{x^2}+\frac{3x}{x^2}-\frac{6}{x^2}}[/math]
The main problem is that you divided the numerator by [imath]x^2[/imath] twice. Do you see that?

(I corrected one typo in addition to lots of missing parentheses and carets.)

Arent i suppose to divide x^2 in the numerator, so wouldn't that be x^4 for the square root and x^2 for x

 

[coooool222]

Your work isn't very clear (an image of handwriting would be better), but I think what you mean by it is mostly correct.

I think your work is this: [math]\lim_{x\to+\infty}\frac{\frac{x}{x^2}\sqrt{\frac{x^2}{x^4}+\frac{4}{x^4}}}{\frac{2x^2}{x^2}+\frac{3x}{x^2}-\frac{6}{x^2}}[/math]
The main problem is that you divided the numerator by [imath]x^2[/imath] twice. Do you see that?

(I corrected one typo in addition to lots of missing parentheses and carets.)

Lmao im so dumb, im supposed to cancel x^2 with x thanks

 

[Steven G]

You divided the numerator by x^2*x^2 and the denominator by x^2!!
If you had 8*4/2 you do NOT compute (8/2)*(4/2)! You only divide one factor by 2! This is what you did in your work!

 

[Dr.Peterson]

Arent i suppose to divide x^2 in the numerator, so wouldn't that be x^4 for the square root and x^2 for x

Lmao im so dumb, im supposed to cancel x^2 with x thanks

I see this kind of mistake a lot: You are accustomed to distributing over a sum, which in this case means dividing each term (as you did in the denominator); and you accidentally do the same with a product, dividing each factor (as you did in the numerator, wrongly).

As a result, you divided by [imath]x^2[/imath] twice (the x, and the radical), rather than dividing just once (or, more simply, dividing the x and the radical each just by x):
[math]\lim_{x\to+\infty}\frac{\frac{x}{x}\sqrt{\frac{x^2}{x^2}+\frac{4}{x^2}}}{\frac{2x^2}{x^2}+\frac{3x}{x^2}-\frac{6}{x^2}}=\lim_{x\to+\infty}\frac{1\sqrt{1+\frac{4}{x^2}}}{2+\frac{3x}{x^2}-\frac{6}{x^2}}=\frac{1\sqrt{1}}{2+0-0}=\frac{1}{2}[/math]