Question

Solve lim [ sqrt2 - 2cosx ] / sin( x- pi/4 ) . x-> pi/4

Answer 1

do you get
\[\frac{0}{0}\]?

Answer 2

Yes

Answer 3

then use l'hopital by taking derivative of numerator and denominator seperately

Answer 4

get
\[\frac{2\sin(x)}{\cos(x-\frac{\pi}{4})}\]

Answer 5

replace x by
\[\frac{\pi}{4}\] to get your answer

Answer 6

Is there a way to do it without ll'hospital's rule? Cause i don' really know how to use the rule .

Answer 7

Or could you show steps?

Answer 8

i can't think of a way without l'hopital.
the derivative of
\[\sqrt{2}-2\cos(x)\] is
\[2\sin(x)\] and the derivative of
\[\sin(x-\frac{\pi}{4})\] is
\[\cos(x-\frac{\pi}{4})\]

Answer 9

so i took the derivative top and bottom separately (not quotient rule) and then compute the limit by substitution.

Answer 10

replace x by
\[\frac{\pi}{4}\] in your new ratio and get
\[\sqrt{2}\]as your limit

Answer 11

Okay , thanks .