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Jonask Group TitleBest ResponseYou've already chosen the best response.1
\[\lim_{x \rightarrow 0}\frac{ \sin(e^{x^2}1) }{ x^2 }\]
 one year ago

nitz Group TitleBest ResponseYou've already chosen the best response.0
try out l hospitals rule
 one year ago

DJVJ Group TitleBest ResponseYou've already chosen the best response.0
you can find answers in algebra books..
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
multiply divide by e^(x^2) 1 to directly reach the ans..
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
as we know these things : 1) e^(x^2) 1 > 0 ,for x>0 2)e^x  1 /x = 1 for x>0 3) sin x / x = 1 for x>0 hope that;d help..
 one year ago

jishan Group TitleBest ResponseYou've already chosen the best response.0
simply apply L HOSPITAL rule.
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.1
it says without the rule,i did this \[\frac{\sin(x^2(\frac{ e^{x^2}1 }{ x^2 }))}{x^2}\] since\[\lim_{h \rightarrow 0} \frac{e^h1}{h}=1\]\[\frac{\sin (x^2(1))}{x^2}=1\]
 one year ago

jishan Group TitleBest ResponseYou've already chosen the best response.0
same result
 one year ago
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