1. ## Limit Help

Limit homework check

I kinda get this but want to make sure my answers are right

59. Suppose limit x->c f(x)=5 and limx->c g(x)=2. Find
a. lim x->c f(x)g(x) b. lim x->c 2f(x)g(x)

for my answers I got 10 for a and 20 for b

60. Suppose lim x->4 f(x)=0 and lim x->4 g(x)=3. Find
a. lim x->4 (g(x)+3) b. lim x->4 xf(x)
c. lim x->4 g^2(x) d. lim x->4 g(x)/f(x)-1

I got 6 for a, 0 for b, 9 for c and -3 for d

61. Suppose lim x->b f(x)=7 and lim x->b g(x)=-3. Find
a. lim x->b (f(x)+g(x)) b. lim ->b f(x) * g(x)
c. lim x->b 4g(x) d. lim x->b f(x)/g(x)

I got 4 for a, -21 for b, -12 for c and -7/3 for d

62. Suppose lim x->-2 p(x)=4, lim x->-2 r(x)=0, and lim x->-2 s(x)=-3. Find
a. lim x->-2 (p(x)+r(x)+s(x)) b. lim x->-2 p(x)*r(x)*s(x)

i got 1 for a and 0 for b
__________________________________________________ _____________
also 45. let f(x)={a-x^2 if x<2
x^2+5x-3 if x is greater than or equal to 2
For what values of a does lim x->-2 f(x) exist

i got 15 but im not sure if thats right

2. ## Re: Limit Help

Originally Posted by Kathrynm77
Limit homework check

I kinda get this but want to make sure my answers are right

59. Suppose limit x->c f(x)=5 and limx->c g(x)=2. Find
a. lim x->c f(x)g(x) b. lim x->c 2f(x)g(x)

for my answers I got 10 for a and 20 for b
Yes, that's right- $\displaystyle \lim_{x\to a} fg(x)= (\lim_{x\to a} f(x))(\lim_{x\to a} g(x))$ and $\displaystyle \lim_{x\to a}(cf)(x)= c\lim_{x\to a} f(x)$.

60. Suppose lim x->4 f(x)=0 and lim x->4 g(x)=3. Find
a. lim x->4 (g(x)+3) b. lim x->4 xf(x)
c. lim x->4 g^2(x) d. lim x->4 g(x)/f(x)-1

I got 6 for a, 0 for b, 9 for c and -3 for d
Yes, for a, b, and c. Is d $\displaystyle \lim_{x\to 4}\frac{g(x)}{f(x)}- 1$ or $\displaystyle \lim_{x\to a}\frac{g(x)}{f(x)- 1}$? The first does not exist. The second is -3.

61. Suppose lim x->b f(x)=7 and lim x->b g(x)=-3. Find
a. lim x->b (f(x)+g(x)) b. lim ->b f(x) * g(x)
c. lim x->b 4g(x) d. lim x->b f(x)/g(x)

I got 4 for a, -21 for b, -12 for c and -7/3 for d
Yes, these are all pretty direct applications of the "limit theorems".

62. Suppose lim x->-2 p(x)=4, lim x->-2 r(x)=0, and lim x->-2 s(x)=-3. Find
a. lim x->-2 (p(x)+r(x)+s(x)) b. lim x->-2 p(x)*r(x)*s(x)

i got 1 for a and 0 for b
Yes.

__________________________________________________ _____________
also 45. let f(x)={a-x^2 if x<2
x^2+5x-3 if x is greater than or equal to 2
For what values of a does lim x->-2 f(x) exist
You mean x->2, not -2. You are given no information about what happens at x=-2.

i got 15 but im not sure if thats right
The "limit from below" is $\displaystyle \lim_{x\to 2} a- x^2= a- 4$. The "limit from above" is $\displaystyle \lim_{x\to 2} x^2+ 5x- 3= 4+ 10- 3= 11$. The limit, itself, exists if and only if those two one-sided limits are the same so we must have a- 4= 11 which does, in fact, give a= 15.

Well done!

3. ## Re: Limit Help

thanks so much!!!