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Jonask

  • 3 years ago

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  1. Jonask
    • 3 years ago
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    \[\lim_{x \rightarrow 0}\frac{ \sin(e^{x^2}-1) }{ x^2 }\]

  2. nitz
    • 3 years ago
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    try out l hospitals rule

  3. nitz
    • 3 years ago
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    @Jonask

  4. DJVJ
    • 3 years ago
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    you can find answers in algebra books..

  5. shubhamsrg
    • 3 years ago
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    multiply divide by e^(x^2) -1 to directly reach the ans..

  6. shubhamsrg
    • 3 years ago
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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..

  7. jishan
    • 3 years ago
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    simply apply L HOSPITAL rule.

  8. Jonask
    • 3 years ago
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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\[\lim_{h \rightarrow 0} \frac{e^h-1}{h}=1\]\[\frac{\sin (x^2(1))}{x^2}=1\]

  9. jishan
    • 3 years ago
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    same result

  10. Jonask
    • 3 years ago
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    as what

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