so far i can only do some types but when they vary its confusing!

lim x^2 − 5x + 2.

x->4

lim5 sin x

x->0 x

lim1

1 + sin x

x->0

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- Jul 8th 2008, 02:50 AMEmmetti need help with the following limits questions
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 - Jul 8th 2008, 03:06 AMChop Suey
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 $ - Jul 8th 2008, 03:15 AMIsomorphismQuote:

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. - Jul 8th 2008, 03:33 AMEmmett
- Jul 8th 2008, 03:38 AMChop Suey
When you plug in 0 into the x, and you don't get an indeterminate form, you would get the limit (1).

Sin(x) equals 0 if x equals 0. - Jul 8th 2008, 04:07 AMEmmett