# i need help with the following limits questions

• July 8th 2008, 02:50 AM
Emmett
i 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
• July 8th 2008, 03:06 AM
Chop Suey
1. $P(x) = x^2 - 5x + 2$

$\lim_{x \to a} P(x) = P(a)$

Only for Polynomials!

2. $f(x) = \frac{5\sin{x}}{x}$

$\lim_{x \to 0} f(x) = \frac{0}{0}$

The limit has the indeterminate form of 0/0, so we use L'Hôpital's rule and differentiate the numerator and denominator.

$\lim_{x \to 0} \frac{5\sin{x}}{x}$

$\lim_{x \to 0} \frac{5\cos{x}}{1} = 5\cos{0} = 5$
• July 8th 2008, 03:15 AM
Isomorphism
Quote:

1. P(x) = x^2 - 5x + 2

Lim P(x) = P(a)
x -> a

Only for Polynomials!
It works for any function continuous at a... after all that is the definition of continuity.

Quote:

Originally Posted by Emmett
lim 1
1 + sin x
x->0

$\lim_{x \to 0} \frac1{1+\sin x} = \frac1{1+0} = 1$

If you believe that $\frac1{1+\sin x}$ is continuous around 0, you can substitute.
• July 8th 2008, 03:33 AM
Emmett
Quote:

Originally Posted by Isomorphism
It works for any function continuous at a... after all that is the definition of continuity.

$\lim_{x \to 0} \frac1{1+\sin x} = \frac1{1+0} = 1$

If you believe that $\frac1{1+\sin x}$ is continuous around 0, you can substitute.

thats really helpful, thanks, just one quick question though, do we take it that sin x is always equal to 0 or was there something else i missed?
thanks again
• July 8th 2008, 03:38 AM
Chop 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.
• July 8th 2008, 04:07 AM
Emmett
Quote:

Originally Posted by Chop 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.

oh rite i get it now, thanks dude