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

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

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

shubhamsrgBest 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

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

JonaskBest 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
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