Question

Does anyone know how to find the limit of this function?

Answer 1

Substituting only gives you \(\dfrac00\)

I would multiply by its conjugate then substitute

Answer 2

\(\color{#0cbb34}{\text{Originally Posted by}}\) @dude
Substituting only gives you \(\dfrac00\)

I would multiply by its conjugate then substitute
\(\color{#0cbb34}{\text{End of Quote}}\)
I tried doing that, but I got
\[\frac{ 1 }{ \sqrt{6y}-\sqrt{7x+1} }\]
And when I subbed in the values, it just doesn't give me a value. Did I do something wrong here?

Answer 3

\(\color{#0cbb34}{\text{Originally Posted by}}\) InsatiableSuffering
\(\color{#0cbb34}{\text{Originally Posted by}}\) dude
Substituting only gives you \(\dfrac00\)

I would multiply by its conjugate then substitute
\(\color{#0cbb34}{\text{End of Quote}}\)
I tried doing that, but I got
\[\frac{ 1 }{ \sqrt{6y}-\sqrt{7x+1} }\]
And when I subbed in the values, it just doesn't give me a value. Did I do something wrong here?
\(\color{#0cbb34}{\text{End of Quote}}\)


\(\dfrac{1}{\sqrt{6y}\color{red}{+}\sqrt{7x+1}}\)

Answer 4

\(\color{#0cbb34}{\text{Originally Posted by}}\) @dude
\(\color{#0cbb34}{\text{Originally Posted by}}\) InsatiableSuffering
\(\color{#0cbb34}{\text{Originally Posted by}}\) dude
Omg thank you so much! That makes so much more sense.
Substituting only gives you \(\dfrac00\)

I would multiply by its conjugate then substitute
\(\color{#0cbb34}{\text{End of Quote}}\)
I tried doing that, but I got
\[\frac{ 1 }{ \sqrt{6y}-\sqrt{7x+1} }\]
And when I subbed in the values, it just doesn't give me a value. Did I do something wrong here?
\(\color{#0cbb34}{\text{End of Quote}}\)


\(\dfrac{1}{\sqrt{6y}\color{red}{+}\sqrt{7x+1}}\)
\(\color{#0cbb34}{\text{End of Quote}}\)

Answer 5

so there is this question to find the limit of this fraction :

\[\frac{ \sqrt{6y -\sqrt{7x+1}} }{ 6y -7x -1 }\]


so given that we need find the limit of this fraction for x=5 and y=6
but first of all if you check these value of x and y substituting inside this faction you get the denominator equal zero what mean that this fraction for x=5 and y=6 is undefined bc. hope you know that a fraction need be defined to can work ,make calcule with ...

so in this way thats all what you can make - from this result DNE - does not exist

hope helped understandably easy

@dude

Answer 6

and any idea about conjugate

so bc. in denominator there isnt radical we dont use conjugate - not is necessary - we use conjugate in case when there is radical in denominator and using conjugate we try eliminate the radical(s) from denominator - hope helped

Answer 7

@AZ

Answer 8

I will have to agree with @dude. When you plug in (5, 6) , you get \(\dfrac{0}{0}\) which is an indeterminate form.

This doesn't mean that the final answer is DNE (which as you can see from the screenshot was incorrect). I do see where Jhonyy is coming from- that conjugates are normally used when we have a radical in the denominator and we're trying to remove it from the denominator.

But it is also a perfectly valid method to multiply the numerator and denominator by the conjugate to see if you can then evaluate the limit. (It's basically like multiplying by 1 so you're not changing anything.)

Also, since OP got the conjugate wrong I'd just like to remind that
The conjugate of a+b is a-b
The conjugate of a-b is a+b

You're essentially changing the signs.
Which reminds me that we also use conjugates to simplify complex numbers like 1/(3+4i)

Answer 9

We can't use L'Hopital's rule since we're dealing with a multivariable function.

As a refresher, L'Hopital's rule allows us to evaluate limits that are indeterminate so either \(\dfrac{0}{0}\) or \(\dfrac{\infty}{\infty}\) and states

\[\lim_{x \rightarrow c} \frac{ f(x) }{ g(x) } = \lim_{x \rightarrow c}\frac{ f'(x) }{ g'(x) }\]

Even if you are taking multivariable calculus, I don't think you would be expected to know nor will they teach you how to use L'Hopital's rule. However, if you are interested:
https://arxiv.org/pdf/1209.0363.pdf

Answer 10

\(\color{#0cbb34}{\text{Originally Posted by}}\) @AZ
We can't use L'Hopital's rule since we're dealing with a multivariable function.

As a refresher, L'Hopital's rule allows us to evaluate limits that are indeterminate so either \(\dfrac{0}{0}\) or \(\dfrac{\infty}{\infty}\) and states

\[\lim_{x \rightarrow c} \frac{ f(x) }{ g(x) } = \lim_{x \rightarrow c}\frac{ f'(x) }{ g'(x) }\]

Even if you are taking multivariable calculus, I don't think you would be expected to know nor will they teach you how to use L'Hopital's rule. However, if you are interested:
https://arxiv.org/pdf/1209.0363.pdf
\(\color{#0cbb34}{\text{End of Quote}}\)
ty @AZ but any idea - opinion about these my words please - :
,,but first of all if you check these value of x and y substituting inside this faction you get the denominator equal zero what mean that this fraction for x=5 and y=6 is undefined bc. hope you know that a fraction need be defined to can work ,make calcule with ..."

Answer 11

Yes, I understand. Because the fraction is undefined we need to multiply by the conjugate so that it won't be undefined at x=5 and y=6.

Answer 12

so sorry but i think we dont understand - i mean that when for any variable values we get for a fraction the denominator equal zero so this mean that this fraction is undefined and for next steps we not can work - make any calcule with this fraction bc this not can be used - how i ve read somewhere : the presently mathematics not is on this level to can make calcules with undefined fractions - so with denominator equal zero ...

Answer 13

Just because the denominator equals 0 does not mean that that the limit is undefined :)

For example

\[\lim_{x \rightarrow 1}\dfrac{x^2 +x-2}{x^2-1}\]

You will get 0/0 if you plug in x=1. But if you graph it, you will notice that the limit approaches 1.5 as x gets closer to 1. You could simplify it and then evaluate the limit. We could have also used L'Hopital's rule to help us in evaluating this limits as well as limits that we would otherwise believe are undefined but are actually not.

But for our original question, multiplying by the conjugate is the best way to go.

Answer 14

\(\color{#0cbb34}{\text{Originally Posted by}}\) @AZ
Just because the denominator equals 0 does not mean that that the limit is undefined :)

For example

\[\lim_{x \rightarrow 1}\dfrac{x^2 +x-2}{x^2-1}\]

You will get 0/0 if you plug in x=1. But if you graph it, you will notice that the limit approaches 1.5 as x gets closer to 1. You could simplify it and then evaluate the limit. We could have also used L'Hopital's rule to help us in evaluating this limits as well as limits that we would otherwise believe are undefined but are actually not.

But for our original question, multiplying by the conjugate is the best way to go.
\(\color{#0cbb34}{\text{End of Quote}}\)
ok ty so much