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))
Hello, homeylova223!
By now, you should know that more parenthese are needed . . .
$\displaystyle \text{(1) as }x\text{ approaches zero: }\:Sin(x)^2/x$
Is that: .$\displaystyle \frac{\sin x^2}{x}\;\text{ or }\;\frac{\sin^2\!x}{x}\;?$
$\displaystyle \text{(2) As }x\text{ approaches zero: }\:Sec(x)-1/x sec(x)$
Is that: .$\displaystyle \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}\;? $
For Q.1, write it as $\displaystyle \displaystyle \lim_{x \to 0}\frac{\sin{x}}{x}\cdot \lim_{x \to 0}\sin{x}$. The first is a standard limit, the second can be evaluated using direct substitution.
For Q.2
$\displaystyle \displaystyle \begin{align*}\lim_{x \to 0}\frac{\sec{x} - 1}{x\sec{x}} &= \lim_{x \to 0}\frac{\frac{1-\cos{x}}{\cos{x}}}{\frac{x}{\cos{x}}} \\ &= \lim_{x \to 0}\frac{1 - \cos{x}}{x}\end{align*}$
This is another standard limit, but if you don't know it, try multiplying top and bottom by the top's conjugate.