Question

limits and continuity see attachment

Answer 1

show pls

Answer 2

you can either use l'hopital for the first one, as suggested by the second problem, or you can factor the denominator as
\[(2x+3)(x-1)=(2x+3)(\sqrt{x}-1)(\sqrt{x}+1)\] and then cancel a factor with the numerator

Answer 3

i cannot use l hospital

Answer 4

show pls

Answer 5

then use what i wrote above. you get
\[\lim_{x\rightarrow 1}\frac{2x-3}{(2x+3)(\sqrt{x}+1)}\] replace x by 1 to get your answer

Answer 6

what about scind one??

Answer 7

..............

Answer 8

hmm i am thinking about how to do the without l'hopital. with l'hopital you get
\[\lim_{x\rightarrow 3} \frac{f'(x)\sqrt{x}}{\sqrt{f(x)}}\]

Answer 9

use the fact that

\[f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}\]

Answer 10

factor the top and bottom

Answer 11

soooooooooo

Answer 12

..........

Answer 13

I gave you a hint...use it

Answer 14

k

Answer 15

next one

Answer 16

...........

Answer 17

aravind use rationalization of numerator

Answer 18

:(

Answer 19

i see nothing that i could offer that hasnt already been presented.

Answer 20

the lim as x->0 of 1+tan(x) = 1 since there is no restrictions for tan(0)

the lim as x->0 of 1+sin(x) = 1 since there is no retrictions to sin(0) and it doesnt make the denom go bad

Answer 21

the only thing to check would be csc(x)

Answer 22

which i believe is the flip of sin(0)

Answer 23

1/sin(0) is bad so you would have to account for it i spose

Answer 24

i see no way that the limit can be defined since it does -inf and +inf from different directions

Answer 25

or am i reading the notation wrong? is that an exponent?

Answer 26

yes

Answer 27

the limit is 1, but that is hard to show without using L'Hospital's rule