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1.lim(x->0)(3x^2+6x)/sin(3x)

=lim(x->0)(3x^2+6x)'/sin(3x)'.....羅必達定理

=lim(x->0)(6x+6)/3cos(3x)

=6/3

=2

2.lim(x->0)5x^2/(sec^2(x)-1)

=lim(x->0)(5x^2)'/(tan^2(x))'

=lim(x->0)10x/2tan(x)sec^2(x)

=lim(x->0)(5xcos^3(x))'/sin(x)'

=lim(x->0)5(cos^3(x)-3xcos^2(x)sin(x))/cos(x)

=5(1-3*0)/1

=5

3.lim(x->-2)[√(13-x^2)-3]/(x+2)

=lim(x->-2)[√(13-x^2)-3]'/(x+2)'

=lim(x->-2)[-x/√(13-x^2)]/1

=-2/√(13-4)

=-2/3

4.lim(x->2)[1/(x+5)-1/(2x+3)]/(x-2)

=lim(x->2)[(2x+3-x-5)/(x+5)(2x+3)]/(x-2)

=lim(x->2)[(x-2)/(x+5)(2x+3)]/(x-2)

=lim(x->2)[1/(x+5)(2x+3)]

=1/7*7

=1/49

5.lim(x->π/4)(tan(x)-1)/(cos(x)-sin(x))

=lim(x->π/4)(tan(x)-1)'/(cos(x)-sin(x))'

=lim(x->π/4) sec^2(x)/(-sin(x)-cos(x))

=2/(-√2)

=-√2

6.lim(x->1)(|x+2|-|x-4|)/(1-x^2)

=lim(x->1)(|x+2|-|x-4|)'/(1-x^2)'

=lim(x->1)(|1|-|1|)/(-2x)

=0/(-2)

=0

7.lim(x->0)x^2*sin(1/x)

=lim(x->0)sin(1/x)/(1/x^2)

=lim(y->∞)sin(y)/y^2.....y=1/x

=(-1~1)/y^2.....-1<=sin(y)<=1

=0

8.lim(x->3)[√(3x-1)-√(2x+2)]/(x-3)

=lim(x->3)[√(3x-1)-√(2x+2)]'/(x-3)'

=lim(x->3)[3/2√(3x-1)-1/√(2x+2)]/1

=3/2√(3x-1)-1/√(2x+2)

=3/2√8-1/√8

=1/2√8

=1/4√2

=√2/8

9a.使用畢氏定理:

Q=atan(3/4) => cos(Q)=4/5.....ans

9b.使用畢氏定理:

A=asin(-12/13) => 3Q or 4Q

=> cos(A)=5/13(4Q) or -5/13(3Q)

10.ln[(x^4+3)^2/x^5*√(x^2+1)]

=ln[(x^4+3)^2]-ln[x^5*√(x^2+1)]

=2*ln(x^4+3)-ln(x^5)-ln(x^2+1)^0.5

=2*ln(x^4+3)-5*ln(x)-0.5*ln(x^2+1)

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• Massacres That Matter - Part 1 - 'Responsibility To Protect' In Egypt, Libya And Syria
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This entry passed through the Full-Text RSS service — if this is your content and you're reading it on someone else's site, please read the FAQ at fivefilters.org/content-only/faq.php#publishers. Five Filters recommends:

• Massacres That Matter - Part 1 - 'Responsibility To Protect' In Egypt, Libya And Syria
• Massacres That Matter - Part 2 - The Media Response On Egypt, Libya And Syria
• National demonstration: No attack on Syria - Saturday 31 August, 12 noon, Temple Place, London, UK