Question

can you help finding this limit please

Answer 1

do u know L'hospitals law

Answer 2

hmmm not really, im not allowed to use l,hospital rule :S

Answer 3

limits or derivatives

Answer 4

\[\lim_{x \rightarrow a}\frac{ \sec(x) -\sec(a)}{ x-a }\]
just diffrentiate denominator and numerator once
\[\lim_{x \rightarrow a}\frac{ \sec(x) \tan(x)-0}{ 1-0 }=\lim_{x \rightarrow a}\sec(x)\tan(x)\\=\sec(a)\tan(a)\]

Answer 5

hmm the derivative of sec x is secxtanx and the derivative of sec a is 0 ?

Answer 6

yeah

Answer 7

hmmm

Answer 8

\[\lim_{x \rightarrow a}\frac{ \sec(x) -\sec(a)}{ x-a }\]

this is the definition of the derivative of the secant evaluated at a. No need to use L'Hospitals rule.

Answer 9

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

Answer 10

so it's like finding the derivative of sec by using that limit definition?

Answer 11

oh nice, ok ill try to work it out myself and come back later thanks everyone :D

Answer 12

yw!!

Answer 13

you don't need to use the limit definition; you just need to recognize that it is the limit definition of a derivative and then use what you know about the derivative to answer the question.

\[\lim_{x \rightarrow a}\frac{ \sec(x) -\sec(a)}{ x-a }=\left.\frac{d}{dx}[\sec(x)]\right|_{x=a}=\sec(a)\tan(a)\]