Question

Evaluate the limit, if it exists. (If an answer does not exist, enter DNE.)
lim
h → 0

(49 + h)^.5 − 7) / h

Answer 1

i get 98+h over h^2

Answer 2

wow a decimal exponent!!

Answer 3

\[\frac{\sqrt{49+h}-7}{h}\] is what you really have
the gimmick here is to rationalize the numerator by multiplying by \[\frac{\sqrt{49+h}+7}{\sqrt{49+h}+7}\]

Answer 4

leave the denominator in factored form, i.e. don't multiply out

Answer 5

wanna try it or you still need help?

Answer 6

i think I got it. Thanks!

Answer 7

yw
let me know what you get and i will check it
btw don't be too confused by these
in a week you will be able to do this problem instantly in your head using a shortcut method

Answer 8

??? whats the method?

Answer 9

it is a secret i can't tell you

Answer 10

do i count \[(\sqrt{49+h}-7)\times(\sqrt{49-h}+7 )\] as 3 variables?

Answer 11

times 3 variables

Answer 12

ok in a week or two you will recognize this as the "derivative" of \(f(x)=\sqrt{x}\) evaluated at \(x=49\)
NO NO NO hold on

Answer 13

if i wrote \(\sqrt{49-h}\) it was a typo
it should be
\[\frac{\sqrt{49+h}-7}{h}\times \frac{\sqrt{49+h}+7}{\sqrt{49+h}+7}\]

Answer 14

the numerator uses \((a-b)(a+b)=a^2-b^2\) which is what makes this trick work

Answer 15

okay so I get 1 over (49+h)^.5 +7

Answer 16

you should get
\[49+h-49=h\] in the numerator

Answer 17

limit is 1/14?

Answer 18

then dividing by \(h\) gives you
\[\frac{1}{\sqrt{49+h}+7}\]
yes

Answer 19

i mean "yes \(\frac{1}{14}\) is correct"

Answer 20

are there online videos that can teach me calculus?

Answer 21

for free?

Answer 22

sure

Answer 23

give me a second and i can link to one or two

Answer 24

thanks!

Answer 25

http://www.hippocampus.org/HippoCampus/Calculus%20%26%20Advanced%20Math

http://patrickjmt.com/

Answer 26

first one is not too dynamic, but decent
second one you pick the video and the guy shows real clearly and simply how to solve a problem
but like i said, don't get too bogged down with these, you will leave this method behind very rapidly

Answer 27

\[(\frac{ 1}{ 9 }+\frac{ 1 }{ x })/9+x\]

Answer 28

limit is dne?

Answer 29

oh no probably not

Answer 30

i take it it is
\[\lim_{x\to 0}\frac{\frac{1}{9}+\frac{1}{x}}{9+x}\] is that right? or with a minus sign?

Answer 31

gotta add up in the numerator

Answer 32

oh I got it!

Answer 33

1/-81

Answer 34

you get
\[\frac{\frac{x+9}{9x}}{9+x}\]

Answer 35

i think just \(\frac{1}{81}\) if you do not have a minus sign there

Answer 36

oh wait, was it the limit as \(x\to -9\)
then you are correct