standard limit indication

V

Vali

Junior Member
Joined
Feb 27, 2018
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87
Hi!
I have the limit from the below picture.
I have to use that formula but I don't know how.
I need to split that limit in 2 limits then the result should give n^2 + n^2 (from that standard limit) and the result will be sqrt(2)^2 + sqrt(2)^2 = 4
Thanks!
 

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Vali said:
Hi!
I have the limit from the below picture.
I have to use that formula but I don't know how.
I need to split that limit in 2 limits then the result should give n^2 + n^2 (from that standard limit) and the result will be sqrt(2)^2 + sqrt(2)^2 = 4
Thanks!
Click to expand...
What formula? You need to understand better what an equal sign means. One of its meaning is that you can replace what is on one side of the equal sign with what is on the other side of the equal sign, You are being asked to replace x with 1/t and sqrt(2) with n. Can you please do that and see if you can go further? Then report back to us showing your work so we can assist you further.
 
tumblr_messaging_pkyw4fAgEW1tyt8xs_1280.jpg How to continue ?
If I add subtract 1 at numerator I can use that standard limit but I don't get the right answer.
 
Vali said:
Hi!
I have the limit from the below picture.
I have to use that formula but I don't know how.
I need to split that limit in 2 limits then the result should give n^2 + n^2 (from that standard limit) and the result will be sqrt(2)^2 + sqrt(2)^2 = 4
Click to expand...
I was able to do this after understanding your notation.
\(\displaystyle \mathop{\lim }\limits_{x \to \infty } \left[ {\dfrac{{{{(x + \sqrt 2)}^{\sqrt 2 }} - {{(x - \sqrt 2 )}^{\sqrt 2 }}}}{{{x^{\sqrt 2 -1}}}}} \right] = \mathop {\lim }\limits_{x \to \infty } \left\{{\dfrac{{{{\left( {1 + \dfrac{{\sqrt 2 }}{x}} \right)}^{\sqrt 2 }} -{{\left( {1 - \dfrac{{\sqrt 2 }}{x}} \right)}^{\sqrt 2 }}}}{{{x^{ -1}}}}} \right\}\)

If \(\displaystyle t=x^{-1}\) then \(\displaystyle \mathop {\lim }\limits_{x \to \infty } \left\{ {\frac{{{{\left( {1 + \dfrac{{\sqrt 2 }}{x}} \right)}^{\sqrt 2 }} - {{\left( {1 - \dfrac{{\sqrt 2 }}{x}} \right)}^{\sqrt 2 }}}}{{{x^{ - 1}}}}} \right\} = \mathop {\lim }\limits_{t \to 0} \left\{ {\dfrac{{{{\left( {1 + \sqrt 2 t} \right)}^{\sqrt 2 }} - {{\left( {1 - \sqrt 2 t} \right)}^{\sqrt 2 }}}}{t}} \right\}\) Now we have \(\displaystyle \dfrac{0}{0}\)

Using l'Hopital's theorem: \(\displaystyle \mathop {\lim }\limits_{t \to 0} \left\{ {\dfrac{{{{\left( {1 + \sqrt 2 t} \right)}^{\sqrt 2 }} - {{\left( {1 - \sqrt 2 t} \right)}^{\sqrt 2 }}}}{t}} \right\}\overbrace {\; = \;}^H\mathop {\lim }\limits_{t \to 0} \left\{ {\dfrac{{\sqrt 2 {{\left( {1 + \sqrt 2 t} \right)}^{\sqrt 2 - 1}}\left( {\sqrt 2 } \right) - \sqrt 2 {{\left( {1 - \sqrt 2 t} \right)}^{\sqrt 2 - 1}}\left( { - \sqrt 2 } \right)}}{1}} \right\} \to 4\)
 
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