$$\lim_{ｘ \rightarrow ０}　\frac{ ｅ^{ｘ^{２}}－１ }{ ２ｘ^{２} }$$　ｆｉｎｄ　ｔｈｅ　ｌｉｍｉｔ　ａｓ　ｘ　ａｐｒｏａｃｈｅｓ　ｚｅｒｏ

Question

anonymous · Answer

are you familiar with l'Hopital's Rule?

zzr0ck3r · Answer

do you know l'hopitals rule?

anonymous · Answer

ｙｅｓ　０／０　ｏｒ　１／１　ｒｉｇｈｔ？

zzr0ck3r · Answer

do it twice

anonymous · Answer

0/0 or inf/inf. 1/1 is fine, limit exists and = 1

anonymous · Answer

２－１／４　＝　１／４

zzr0ck3r · Answer

?

anonymous · Answer

；（　ｔｈａｔ＇ｓ　ｎｏｔ　ｉｔ

zzr0ck3r · Answer

you will get (4x^2*e^x^2)/4  --> 0

anonymous · Answer

ｏｈ　ｔｙ

anonymous · Answer

i get limit = 1/2

anonymous · Answer

１／２，　ｏｒ　ｚｅｒｏ

zzr0ck3r · Answer

second derivative of the top is 4x^2 * e^x^2 right?

zzr0ck3r · Answer

second on bottom is 4

zzr0ck3r · Answer

o im out of it, I didn't use product rule...

zzr0ck3r · Answer

ignore me please

zzr0ck3r · Answer

I don't have paper and im lazy:)

anonymous · Answer

first time
$$\frac{ 2xe^{x^{2}} }{ 4x } $$

second time

$$\frac{ 2e^{x^{2}} + 4x^{2}e^{x^{2}} }{ 4 }$$

anonymous · Answer

yup. only thing was product rule

anonymous · Answer

ｃａｎ　ｉ　ｓｔａｒｔ　ａｎｏｔｈｅｒ　ｔｈｒｅａｄ　ｓｏ　ｉ　ｃａｎ　ｇｉｖｅ　ｙｏｕ　ａｎｄ　ｍｅｔａｌ　ｔｏｏ　Ｅｕｌｅｒ．　Ｂｅｃａｕｓｅ　ｙｏｕ　ｂｏｔｈ　ａｒｅ　ｓｕｐｅｒ　ｈｅｌｐｆｕｌｌ　：）