Question

is this limit solved right,correct ?

lg mean logarithm base ten

lim (lg 100^(x^2) -(lg 10^(x)) =
x->1

= lim (lg 10^(2x^2) -(lg 10^(x)) =
x->1

= lim (lg (10^(2x^2))/(10^(x)) =
x->1

= lim (lg 10^(2x^2 -x)) =
x->1

= lim ((2x^2 -x)lg 10) =
x->1

= lim (2x^2 -x) = 1
x->1

@TheSmartOne
@ganeshie8
@Kainui
@Nnesha

ty.

Answer 1

@phi please ,do you agree with this ? ty.

Answer 2

Correct, but you could eliminate some steps.
\[\begin{align*}\lim_{x\to1}\left(\log_{10}10^{2x^2}-\log_{10}10^x\right)&=\lim_{x\to1}\left(2x^2\log_{10}10-x\log_{10}10\right)\\[1ex]
&=\lim_{x\to1}\left(2x^2-x\right)\end{align*}\]No need to make use of all those other properties of logarithms, but what you've done is certainly valid.

Answer 3

jhonny why are you level79

Answer 4

i m level 74 than i see it now right

Answer 5

@ageta where you see level 79 ?

Answer 6

http://prntscr.com/cii66v

Answer 7

are you the real jhonyy

Answer 8

yes sure from Central Europe ,country Hungary,city Budapest with birth country Romania,the part of Transylvania

so why ? there are again so much jhonyy9 ?

Answer 9

@ageta but than you check my profile i m level 74 - i dont know where you see this level 79 - ? sorry

Answer 10

and jhonyy has lots of fan and you dont

Answer 11

i m jhonny9

Answer 12

@jhonyy9

check here my profile

Answer 13

yes sure its possible that you remember on me when i was champion with level 99"s but my this profile was deleted like these so much other again - hope you know about this cases from the month of may this year

Answer 14

so and i was necessary rebuild a new profile started from level 15

Answer 15

@jhonyy who are this jhonyy ?

Answer 16

what you posted there on this

ageta Human Calculator Best Response Medals 0

http://prntscr.com/cii66v


so this is my posted photo and profile but level 74

Answer 17

do you know anything about this case - or can i help you ,do you like knowing again about me or about my past profile or ???

Answer 18

@TheSmartOne what you say about this above wrote please ? ty . this mean that there are other jhonny with my profile photo ? i dont believe it

Answer 19

ty.

Answer 20

:D
haha :=))

Answer 21

you forgot to add street #
goin to Europe

Answer 22

really @Nnesha do you like viste me here - ok. i wait you

Answer 23

You can just substitute directly (right away), there is nothing wrong with that.

\(\color{blue }{\displaystyle \lim_{x\to1} (\log [100^{x^2}] -\log [10^x])= }\)

\(\color{blue }{\displaystyle \lim_{x\to1} (\log [100^{\color{red}{1}^2}] -\log [10^\color{red}{1}]) = }\)

\(\color{blue }{\displaystyle \lim_{x\to1} (2 -1)= 1. }\)