Question

how to solve: lim (x-->5) (1/x-1/5)(2/x-5)

Answer 1

I can't tell if you are supposed to have more brackets in there or not.

Answer 2

\[\lim_{x \rightarrow 5} (1/x - 1/5) (2/x-5)\]

Answer 3

@azalea Why don't you just put x=5 and see what you get???

Answer 4

First you take the limit of the function to see if it would have a value or not
(1/5-1/5)(2/5-5)
(0 . infinity)
It is in an indeterminate form so combine the fractions
(10-2x)/(5x^2-25x)
Then take the limit
10-2(5)/5(5^2)-25(5)
=0/0 (Still indeterminate)
Use lhospitals rule
take derivative of numerator / take derivative of denominator
-2/10x-25
Now evaluate the lmit
-2/10(5)-25
-2/50-25
-2/25

Answer 5

@ash2326 because it gives me 0

Answer 6

0 is a correct limit. The only problem is when you get an error (aka an indeterminate form).

Answer 7

i have problems with knowing when it is indeterminate and when to use lhosiptals rule

Answer 8

Ok, so you need to use L'Hopital's rule when you have the form:

\[\frac {0}{0} \text{ or } \frac {\pm \infty}{0} \text{ or } \frac{\pm \infty} {\pm \infty}\]

Answer 9

Those are all the indeterminate forms.

Answer 10

thank you! I'll keep that in mind

Answer 11

Also keep in mind that if you have:
\[\infty \times 0\]
You need to make it into the form:
\[\infty \times 0 = \frac {\infty} {\frac{1}{0}} = \frac {\infty}{\infty}\]
So you are then able to use L'Hopital's rule.

Answer 12

oh ok! NOw i know why i can never solve some problems

Answer 13

That should help a lot then :)

Answer 14

i have another limits problem

Answer 15

\[\lim_{x \rightarrow \infty} x-\sqrt{x ^{2} +2x -4}\]

Answer 16

First step: factor out an x.

Answer 17

@azalea close this question and post a new question. That way you'll get more users to answer

Answer 18

thank you!

Answer 19

Then you use the step that I showed you above ;)