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Jonask
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\[\lim_{x \rightarrow 0}\frac{ \sin(e^{x^2}-1) }{ x^2 }\]
nitz
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try out l hospitals rule
nitz
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@Jonask
DJVJ
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you can find answers in algebra books..
shubhamsrg
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multiply divide by e^(x^2) -1 to directly reach the ans..
shubhamsrg
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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..
jishan
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simply apply L HOSPITAL rule.
Jonask
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it says without the rule,i did this
\[\frac{\sin(x^2(\frac{ e^{x^2}-1 }{ x^2 }))}{x^2}\]
since\[\lim_{h \rightarrow 0} \frac{e^h-1}{h}=1\]\[\frac{\sin (x^2(1))}{x^2}=1\]
jishan
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same result
Jonask
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as what