- #1

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[tex]\lim_{x \rightarrow 0} x = \lim_{x \rightarrow 0} sinx[/tex] and so [tex]\lim_{x \rightarrow 0} \frac{x}{x} = 1 = \lim_{x \rightarrow 0} \frac{sinx}{x}[/tex] ??

- Thread starter JG89
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- #1

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[tex]\lim_{x \rightarrow 0} x = \lim_{x \rightarrow 0} sinx[/tex] and so [tex]\lim_{x \rightarrow 0} \frac{x}{x} = 1 = \lim_{x \rightarrow 0} \frac{sinx}{x}[/tex] ??

- #2

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No! Think about using your method to find

[tex]\lim_{x \to 0} x[/tex] by writing [tex]x=\frac{ x^2 }{ x }[/tex]. Also the limit of a quotient is only the quotient of the limits when the bottom limit is nonzero!

Use L'Hospital!!!

- #3

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Ah! That's the part I forgot! That the limit of a quotient is a quotient of limits only when the denominator isn't 0. I constructed a counter-example too to prove myself wrong in the previous post.

[tex]\lim_{x \to 0} x[/tex] by writing [tex]x=\frac{ x^2 }{ x }[/tex]. Also the limit of a quotient is only the quotient of the limits when the bottom limit is nonzero!

Use L'Hospital!!!

[tex]\lim_{x \rightarrow 0} x = 0 [/tex] and so [tex]\lim_{x \rightarrow 0} \frac{x}{x} = 1 = \lim_{x \rightarrow 0} \frac{0}{x} = 0 [/tex] which obviously isn't true (because the denominator is tending to 0).

Thanks for the reply :)

- #4

lurflurf

Homework Helper

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Use L'Hospital!!!

Why do people want to use L'Hospital on every problem? Is it because it is fun to say? L'Hospital is not needed.

lim sin(x)/x=lim [sin(0+x)-sin(x)]/[(0+x)-0]=sin'(0)=1

We might want to compute the limit directly from the definition. For that we need to choose a definition.

def1:

sin(x+y)=sin(x)cos(y)+cos(x)sin(y)

cos(x+y)=cos(x)cos(y)-sin(x)sin(y)

cos(x)^2+sin(x)^2=1

lim sin(x)/x=0

here the limit is included in the definition

QED

def2:

sin(x)=x-x^3/6+...

we can write

sin(x)=sin(x)+sin'(0) x+sin''(0) x^2/2+sin'''(0) x^3/6+...

equating like terms yields

sin'(0)=1

QED

def3:

sin''(x)+sin(x)=0

sin(0)=0

sin'(0)=1

sin'(0)=1 is in the definition

QED

def4:

sin(x)/x=(1-[x/(1pi)]^2)(1-[x/(2pi)]^2)(1-[x/(3pi)]^2)(1-[x/(4pi)]^2)(1-[x/(5pi)]^2)...

lim sin(x)/x=(1-[0/(1pi)]^2)(1-[0/(2pi)]^2)(1-[0/(3pi)]^2)(1-[0/(4pi)]^2)(1-[0/(5pi)]^2)...

=1

QED

def5:

various geometric junk from which it is noticed

for x small

sin(2x)/(2x)<=cos(x)<=sin(x)/x<=1

from which we conclude

cos(0)<=lim sin(x)/x<=1

lim sin(x)/x=1

QED

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