# Limits involving trigonometric functions

## Homework Statement

1)

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

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

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

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dextercioby
Homework Helper
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.

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...?

dextercioby
Homework Helper
To divide something by x means to multiply it by 1/x. So you realize that

$$\frac{7}{6} = \frac{\frac{7}{2}}{\frac{6}{2}}$$

, right ?

To divide something by x means to multiply it by 1/x. So you realize that

$$\frac{7}{6} = \frac{\frac{7}{2}}{\frac{6}{2}}$$

, 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?

eumyang
Homework Helper
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?
$$\displaystyle\lim_{x \rightarrow 0} \frac{3x}{x} = \displaystyle\lim_{x \rightarrow 0} 3 = ?$$

Limit....does not exist......?

VietDao29
Homework Helper
Limit....does not exist......?
Well, you seem to have misinterpret the word 'limit'. The function $$\frac{3x}{x}$$ 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 $$\frac{3x}{x}$$ has the limit of 3. This is because: $$\frac{3x}{x} = 3, \forall x \neq 0$$, so when x tends to 0, $$\frac{3x}{x} \rightarrow 3$$.

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.