Question

Evaluate the given function..............( see the comments for the question)

Answer 1

here is the question

Answer 2

Please explain how did you arrive at the answer, I don't know how to use L'Hopital's rule, nor has my professor taught us that.

Answer 3

so , I found 1/10

Answer 4

Lemme try.....

Answer 5

Sorry, its the wrong answer

Answer 6

ok let me check my work .)

Answer 7

r u sure it's wrong .s

Answer 8

yeah, i did check it, the question system says, its wrong.

Answer 9

Okay, I double checked, your answer is right, sorry...

Answer 10

How did you get it??

Answer 11

Change y=1/x and y-->0

sqrt(25/y^2 + 1/y) - 5/y=

sqrt[(25 + y)/y^2] - 5/y=[sqrt(25+y) - 5]/y

Let's complete the square from 25+y by +/-(y/10)^2
25+y=25 + 2* 5* y/10 + (y/10)^2 - (y/10)^2=(5+y/10)^2-(y/10)^2

Continue sqrt[(5+y/10-5]/y=1/10

lim(x-->infinity)=1/10

Answer 12

okay why did we do the first step?

Answer 13

@Emineyy, why did u do the first step??

Answer 14

Dude, could u tell me how to approach solving this question??

Answer 15

Wait.... am I missing something?

It's:
\[\Large \lim_{x \rightarrow \infty}\sqrt{25x^2+x}-5x\]
right?

The limit of a square root is the square root of a limit.
The limit of a polynomial as \({x \rightarrow \infty}\) is determined by the behavior of the leading term (sign of coefficient, and degree).

The limit of a difference is the difference of the limits.

I think that's all you need to use here??

Answer 16

yup, you got the question right.

Answer 17

the answer is correct

Answer 18

I checked it, initially the website said, its wrong, but when I double checked it came out to be right

Answer 19

It said 1/10 was the limit?

Answer 20

yes

Answer 21

Oh nvr mind... that's what Wolfram says too. OK, I'm bowing out, I'm obviously missing something here, lol.

Answer 22

:)))

Answer 23

Hey, you are back, help me out here bud

Answer 24

So, I don't get why we did the first step and why we did the last step where 5+y/10-5 = 1/10

Answer 25

@Emineyy

Answer 26

help me out bud....@Emineyy

Answer 27

I mean, isn't y = 1/x and if that's so, then shouldn't it be 0??

Answer 28

Well, I do see why it isn't as simple as I thought initially... duh....

The limit of the square root part is + infinity
The limit of the -5x is - infinity

So it isn't like they can just be added. <feeling silly>

So let me see if I can make sense of her approach... she's making a substitution, re-writing with \(y=1/x\) so then \(x=1/y\) and \(x^2=1/y^2\).

\(\sqrt{25/y^2 + 1/y} - 5/y= \sqrt{(25 + y)/y^2} - 5/y=\dfrac{\sqrt{(25+y)} - 5}{y} \)

=====I don't understand what the purpose of all this completing the square stuff was about...??=====
Let's complete the square from 25+y

\(25+y=25 + 2\cdot5\cdot \dfrac{y}{10} + (\dfrac{y}{10})^2 - (\dfrac{y}{10})^2=(5+y/10)^2-(y/10)^2 \)

OK, so:
\(25+y=(5+y/10)^2-(y/10)^2 \)
=====================================
Now back to the limit... with the substitution, we have:

\(\lim_{y \rightarrow 0}\dfrac{\sqrt{(25+y)} - 5}{y} \) which is a 0/0 form and L'hopital's rule applies.

\(\dfrac{d}{dy}(25+y)^{1/2} - 5=\dfrac{1}{2}(25+y)^{-1/2}=\dfrac{1}{2}\dfrac{1}{(25+y)^{1/2}}\)

so:
\(\lim_{y \rightarrow 0}\dfrac{1}{2}\dfrac{1}{(25+y)^{1/2}}=\dfrac{1}{2\cdot5}=\dfrac{1}{10}\)

(And the derivative of the "y" den'r is just 1).