Question

find the limit lim as x->approaches infinity
-7x/sq rt(25x^2+4)

Answer 1

i have tried x by sqr rt of that thing in the denominator but idk if its right..... :l

Answer 2

It might not be immediately obvious, but the top and bottom x's are of the same DEGREE. There is a little trick to problems like this.

When the degree of the top is equal to the degree of the bottom, as you go towards infinity, they'll approach a value of the coefficients on the leading terms.

In this case, the -7 and the sqrt(25) being the coefficients.
But we'll do the steps to make sure it makes sense and all that :)

I'm not sure what you mean by x by sqr rt thing..?

Answer 3

o that makes sense! never mind my gibberish go on....:)

Answer 4

\[\huge \lim_{x \rightarrow \infty}\frac{-7x}{\sqrt{25x^2+4}}\]So here is our problem :D Ummm how have you been learning these problems? By factoring out a term from the square root, or
By multiplying the top and bottom by a reciprocal of the highest x value?

Maybe that was hard to understand the way I said it :3
I just wanna try to solve it the way you're familiar with.

Answer 5

Typing typing typing :D I woulda thought that was an easy question XD hehe

Answer 6

well honestly i kinda forgot how i did problems just because my WONDERFUL teacher decided to add in a limits problem to the test that we are taking tomorrow..which is something we learned a while ago :( .:( but now thinking about it i think that the first one u said about factoring out a term i think we learned it that way.......o hahaha sorry Zep....:(

Answer 7

Factoring out a term? Ok let's try that, maybe it'll make sense soon enough ^^

Answer 8

So in the bottom, let's factor out an x^2 from each term, giving us,\[\large \lim_{x \rightarrow \infty}\frac{-7x}{\sqrt{x^2(25+\frac{4}{x^2})}} \quad = \quad \huge \lim_{x \rightarrow \infty}\frac{-7x}{x\sqrt{25+\frac{4}{x^2}}}\]

Answer 9

Technically that x value should be |x|, but that will only cause an issue if x is heading towards 0. We don't need to worry about it in this case, we know that it is positive x as x->infty.

Answer 10

Understand what I did there with the factoring? It's a little tricky :D

Answer 11

definitely tricky :( um well honestly i kinda get for the most part...... im kinda lost :( all did was take out an x^2 right? i guess what i dont understand it how did why did 4 become 4/x^2?

Answer 12

Here is how factoring works,

You divide the value OUT of each term,
and save the multiplication for later.
We write the value on the outside, with brackets to show the multiplication.

Here's a quick example.\[2+6x\]If I want to divide 2 out of each term, in 2 steps, we'll divide a 2 out of each term, and save the multiplication for later.\[2\left(\frac{2}{2}+\frac{6x}{2}\right)\]Which of course simplifies to \[\large 2(1+3x)\]

Answer 13

So in our problem, let's ignore the squareroot for a moment.\[\large 25x^2+4\]We'll divide the highest degree of x out of each term. Hmm, that appears to be x to the 2nd degree.\[\large x^2\left(\frac{25x^2}{x^2}+\frac{4}{x^2}\right)\]

Answer 14

wait what did i do????? lol

Answer 15

that was WEIRD okay yeah that makes sense so far :)

Answer 16

Crap i gotta go! :C ill be back in like 30 mins if still around :C

Answer 17

ok then thanks anyways :) good LUCK! :)

Answer 18

i'll c :)

Answer 19

\[\large \large x^2\left(\frac{25x^2}{x^2}+\frac{4}{x^2}\right) = x^2\left(25+\frac{4}{x^2}\right)\]

Answer 20

Pulling the x^2 out of the square root gives us
\[\large \lim_{x \rightarrow \infty}\frac{-7x}{\sqrt{x^2(25+\frac{4}{x^2})}} \quad = \quad \lim_{x \rightarrow \infty}\frac{-7x}{x\sqrt{25+\frac{4}{x^2}}}\]

Answer 21

Then the x's will cancel out.\[\lim_{x \rightarrow \infty}\frac{-7}{\sqrt{25+\frac{4}{x^2}}}\]As x approachs infinity, the fraction will get closer and closer to 0.\[\frac{-7}{\sqrt{25+\frac{1}{\infty}}}\]
\[\large \frac{-7}{\sqrt{25+0}}\]