so far i can only do some types but when they vary its confusing!
lim x^2 − 5x + 2.
x->4
lim 5 sin x
x->0 x
lim 1
1 + sin x
x->0
1. $\displaystyle P(x) = x^2 - 5x + 2$
$\displaystyle \lim_{x \to a} P(x) = P(a)$
Only for Polynomials!
2. $\displaystyle f(x) = \frac{5\sin{x}}{x}$
$\displaystyle \lim_{x \to 0} f(x) = \frac{0}{0}$
The limit has the indeterminate form of 0/0, so we use L'Hôpital's rule and differentiate the numerator and denominator.
$\displaystyle \lim_{x \to 0} \frac{5\sin{x}}{x}$
$\displaystyle \lim_{x \to 0} \frac{5\cos{x}}{1} = 5\cos{0} = 5 $
It works for any function continuous at a... after all that is the definition of continuity.1. P(x) = x^2 - 5x + 2
Lim P(x) = P(a)
x -> a
Only for Polynomials!
$\displaystyle \lim_{x \to 0} \frac1{1+\sin x} = \frac1{1+0} = 1$
If you believe that $\displaystyle \frac1{1+\sin x}$ is continuous around 0, you can substitute.