Limit of tan(x)/x as x approaching zero

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  • #2
Svein
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Hints: [itex]tan(x)=\frac{sin(x)}{cos(x)} [/itex] and cos(0) = 1.
 
  • #3
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Furthermore use sqeeze theorem to calculate [tex]\lim_{x\to 0}{\frac{\sin{x}}{x}}[/tex]
L'Hospital rule will not work in that limit
 
  • #5
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Hints: [itex]tan(x)=\frac{sin(x)}{cos(x)} [/itex] and cos(0) = 1.
I know that I can make it lim x->0 tan(x) / x = lim x->0 1/cos(x) * sin(x)/x = 1 * 1 = 1

But, the hint in http://math.stackexchange.com/quest...-lim-limits-x-to0-frac-tan-xx-1#answer-448210 says

lim x->0 tan(x)/x = lim x->0 tan(x)-tan(0)/ x-0
If I plug x=0, the denominator will be zero.
Then, how to solve it using that hint ?
What trigonometric identity should be used ?
 
  • #6
Svein
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What trigonometric identity should be used ?
Well, using l'Hôpital directly, you get...
 
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  • #7
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Furthermore use sqeeze theorem to calculate [tex]\lim_{x\to 0}{\frac{\sin{x}}{x}}[/tex]
L'Hospital rule will not work in that limit
Sure it will.
##\lim_{x \to 0}\frac{sin(x)}{x} = \lim_{x \to 0}\frac{cos(x)}{1} = 1##
 
  • #8
pasmith
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Sure it will.
##\lim_{x \to 0}\frac{sin(x)}{x} = \lim_{x \to 0}\frac{cos(x)}{1} = 1##
L'Hopital's rule does indeed work here, with the caveat that the argument is circular if you are trying to prove that [itex]\sin'(0) = \lim_{x \to 0} \frac{\sin x}x[/itex] exists and is equal to 1.
 

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