1. ## Trigonometric limits?

How Would I solve these two limit

1. As x approaches zero

Sin(x)^2/x

2. As x approaches zero

Sec(x)-1/x sec(x)

This is what I have done

((1/cos(x)-1))/(x)(1/cos(x))

2. ## Re: Trigonometric limits?

Hello, homeylova223!

By now, you should know that more parenthese are needed . . .

$\text{(1) as }x\text{ approaches zero: }\:Sin(x)^2/x$

Is that: . $\frac{\sin x^2}{x}\;\text{ or }\;\frac{\sin^2\!x}{x}\;?$

$\text{(2) As }x\text{ approaches zero: }\:Sec(x)-1/x sec(x)$

Is that: . $\sec x - \frac{1}{x}\sec x\;\text{ or }\;\sec(x) - \frac{1}{x\sec x}\;\text{ or }\;\frac{\sec x - 1}{x\sec x}\;?$

3. ## Re: Trigonometric limits?

For the both it is the latter

(sin^2(x))/(x) This one I've managed to solve using a graphic calculator but I am wondering how to do it analytically

((secx-1))/((x sec x))

4. ## Re: Trigonometric limits?

Originally Posted by homeylova223
For the both it is the latter

(sin^2(x))/(x) This one I've managed to solve using a graphic calculator but I am wondering how to do it analytically

((secx-1))/((x sec x))
For Q.1, write it as $\displaystyle \lim_{x \to 0}\frac{\sin{x}}{x}\cdot \lim_{x \to 0}\sin{x}$. The first is a standard limit, the second can be evaluated using direct substitution.

For Q.2
\displaystyle \begin{align*}\lim_{x \to 0}\frac{\sec{x} - 1}{x\sec{x}} &= \lim_{x \to 0}\frac{\frac{1-\cos{x}}{\cos{x}}}{\frac{x}{\cos{x}}} \\ &= \lim_{x \to 0}\frac{1 - \cos{x}}{x}\end{align*}

This is another standard limit, but if you don't know it, try multiplying top and bottom by the top's conjugate.

5. ## Re: Trigonometric limits?

Hmm For Q.2 I do not understand how you can go from

((1/cos(x)-1))/(x)(1/cos(x))

to (1-cos(x)/(cos(x)/(x)/(cosx)

6. ## Re: Trigonometric limits?

Originally Posted by Prove It
try multiplying top and bottom by the top's conjugate.
Originally Posted by homeylova223
Hmm For Q.2 I do not understand how you can go from

((1/cos(x)-1))/(x)(1/cos(x))

to (1-cos(x)/(cos(x)/(x)/(cosx)
Taking Prove It's suggestion multiply by the top's conjugate

$\displaystyle \lim_{x \to 0} \dfrac{1-\cos(x)}{x} \cdot \dfrac{1+\cos(x)}{1+\cos(x)} = \lim_{x \to 0} \dfrac{1-\cos^2(x)}{x(1+\cos(x))}$

edit: correction

7. ## Re: Trigonometric limits?

Originally Posted by e^(i*pi)
$\displaystyle \lim_{x \to 0} \dfrac{1-\cos(x)}{x} \cdot \dfrac{1+\cos(x)}{1+\cos(x)} = \lim_{x \to 0} \dfrac{1-\cos^2(x)}{1+\cos(x)}$
Not quite, you forgot a term in the last expression.

8. ## Re: Trigonometric limits?

Originally Posted by homeylova223
Hmm For Q.2 I do not understand how you can go from

((1/cos(x)-1))/(x)(1/cos(x))

to (1-cos(x)/(cos(x)/(x)/(cosx)
Get a common denominator for $\displaystyle \frac{1}{\cos{x}} - 1$

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### find the limit as x approaches 0 of 1-cosx over 1-secx

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