anonymous
  • anonymous
limit
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[\lim_{x \rightarrow 0}\frac{ \sin(e^{x^2}-1) }{ x^2 }\]
anonymous
  • anonymous
try out l hospitals rule
anonymous
  • anonymous
@Jonask

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

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