Difference between revisions of "User talk:Kaisssssgao"
Kaisssssgao (talk | contribs) (→QUESTION) |
Kaisssssgao (talk | contribs) (→Solution 2) |
||
| (19 intermediate revisions by 2 users not shown) | |||
| Line 11: | Line 11: | ||
D)6 | D)6 | ||
| − | E) | + | E)undefined |
| + | |||
| + | =Solution greg346 1= | ||
| + | |||
| + | according to kai's theorem, so we chose (A) | ||
| + | 1+1=3~Tyler Xiang | ||
| + | |||
| + | =Solution 2= | ||
| + | |||
| + | according to L'Hopital rule, that lim x->c f(x)/g(x)=lim x->c f'(x)/g'(x), rewrite 1+1=1+1/1, according to the theorem, set 1=f(X) f'(x)=0, set 1=g(x),so g'(x)=0, so 1+1=undefined we chose (E)~Kai Gao | ||
| + | |||
| + | =Solution 3= | ||
| + | never gonna give u up. | ||
Latest revision as of 02:16, 7 July 2023
QUESTION
WHAT'S THE VALUE OF 1+1?
A)3
B)4
C)5
D)6
E)undefined
Solution greg346 1
according to kai's theorem, so we chose (A) 1+1=3~Tyler Xiang
Solution 2
according to L'Hopital rule, that lim x->c f(x)/g(x)=lim x->c f'(x)/g'(x), rewrite 1+1=1+1/1, according to the theorem, set 1=f(X) f'(x)=0, set 1=g(x),so g'(x)=0, so 1+1=undefined we chose (E)~Kai Gao
Solution 3
never gonna give u up.
