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Limits involving trigonometric functions

  • Thread starter Ironside
  • Start date
  • #1
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Homework Statement



1)

lim sin^2(3x)/x^2
x->0

2) lim 2sin5x/(3x-2tan2x)
x->0

Homework Equations



lim sinx/x = 1
x->0


lim tanx/x = 1
x->0

The Attempt at a Solution



We just began working with limits, and we haven't covered much of trig functions at all, but our prof gave us these 2 questions to see what we can do with them. I really don't know how to begin with them.

For the first problem i tried to answer it by splitting it into 2 fractions: sin3x/x * sin3x/x then sin3x/3x * sin3x/3x * 9/1 which gave me the answer 9 for it. I'm not sure if that's right though.

I have no clue how to start on the second one. :(
 

Answers and Replies

  • #2
dextercioby
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The first one is perfect. For the second, I can hint you towards dividing both the denominator and the numerator by x and using the 2 limits given.
 
  • #3
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The first one is perfect. For the second, I can hint you towards dividing both the denominator and the numerator by x and using the 2 limits given.
I'm not following you, you mean times it by 1/x or....times x/x...?
 
  • #4
dextercioby
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To divide something by x means to multiply it by 1/x. So you realize that

[tex] \frac{7}{6} = \frac{\frac{7}{2}}{\frac{6}{2}} [/tex]

, right ?
 
  • #5
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To divide something by x means to multiply it by 1/x. So you realize that

[tex] \frac{7}{6} = \frac{\frac{7}{2}}{\frac{6}{2}} [/tex]

, right ?
Ok so what i did is (2sin5x/x)/((3x-2tan2x)/x) and i separated it into 2 limits so

lim 2sin5x/x , then i do 2sin5x/5x * 5/1 then i get for the top fraction 10
x->0


then for the (3x-2tan2x)/x , i did 3x/x - 2tan2x/x , then i did 2tan2x/2x * 2/1 which gave m 4, then obviously i run into the problem that lim 3x/x when x ->0 is undefined. What did i do wrong?
 
  • #6
eumyang
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then for the (3x-2tan2x)/x , i did 3x/x - 2tan2x/x , then i did 2tan2x/2x * 2/1 which gave m 4, then obviously i run into the problem that lim 3x/x when x ->0 is undefined. What did i do wrong?
[tex]\displaystyle\lim_{x \rightarrow 0} \frac{3x}{x} = \displaystyle\lim_{x \rightarrow 0} 3 = ?[/tex]
 
  • #7
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Limit....does not exist......?
 
  • #8
VietDao29
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Limit....does not exist......?
Well, you seem to have misinterpret the word 'limit'. The function [tex]\frac{3x}{x}[/tex] is indeed undefined at x = 0.

However, when x tends to 0 (note that: x only tends to 0, i.e x is 'closed enough' to 0, not x = 0), then the function [tex]\frac{3x}{x}[/tex] has the limit of 3. This is because: [tex]\frac{3x}{x} = 3, \forall x \neq 0[/tex], so when x tends to 0, [tex]\frac{3x}{x} \rightarrow 3[/tex].

When talking about taking the limit as x tends to some value a, it means that x is near/close enough to a. The function may be undefined at x = a, but may have limit when x tends to a.
 

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