Question

Find:the limit

Answer 1

\[a.\lim_{x \rightarrow -4} x^2-16 \div 2x^2+8x\]

Answer 2

b. \[\lim_{x \rightarrow 0} \sin5x \div 10x\]

Answer 3

first one
factor numerator and denominator, cancel. then replace x by -4

Answer 4

you will get a common factor of \((x+4\) top and bottom, and after you cancel the fraction will be defined at \(x=-4\)

Answer 5

second one rewrite as
\[\frac{1}{2}\frac{\sin(5x)}{5x}\] the limit as \(x\to 0\) of \(\frac{\sin(x)}{x}=1\) so you get \(\frac{1}{2}\)

Answer 6

or u can use lhopitals rule for the second one :D

Answer 7

so the answer for the first one is 1 right?

Answer 8

@lalaly show me lhopital rule plz

Answer 9

first one is 1 yup and ill show u how to apply lhopitals

Answer 10

ok because the more ways one maths can be solved is better so i can choose the one i understand best

Answer 11

if u substitute 0 in sin5x/10x you get 0/0 which is called indeterminate form

so u should apply lhopitals rule which means u find the derivative of the nominator and then find the derivative of the denominator and then find the llimit of that
\[\lim_{x \rightarrow 0}\frac{5\cos(5x)}{10}\]when u plug 0 in x
u get 5/10 which is 1/2

Answer 12

i am going to guess that you have not gotten to l'hopital's rule, and in fact may have no gotten to derivatives yet, since they are not needed for these problems. if you didn't get there yet, l'hopital's rule will make no sense

Answer 13

@satellite73 i got some rules in my notebook yes but didn't know the name of it

Answer 14

thanks both of you :)