Question

I do not know how to approach these. I can go further in #2 but I have no idea how to even start #1. Both these limits are
limx→Π/4


1.
sinx−cosxtanx−1


2.
(1tanx−1−2tan2x−1)


And for this one I got 4 as the answer but the answer is 2

limx→4x−4x√−8−x−−−−−√

Answer 1

I do not know how to approach these. I can go further in #2 but I have no idea how to even start #1. Both these limits are
\[\lim_{x \rightarrow \frac{ \Pi }{ 4 } }\]

1. \[\frac{ \sin x-\cos x }{ \tan x-1 }\]


2. \[(\frac{ 1 }{ \tan x-1 }-\frac{ 2 }{\tan ^{2}x-1 })\]

Another one which is different And for this one I got 4 as the answer but the answer is 2.

3. \[\lim_{x \rightarrow 4}\frac{ x-4 }{ \sqrt{x}-\sqrt{8-4} }\]

Answer 2

We have
\[\lim_{x\to \frac{\pi}{4}}\frac{ \sin x-\cos x }{ \tan x-1 }\]
We know tan x=sin x \ cos x so
\[\lim_{x\to \frac{\pi}{4}}\frac{ \sin x-\cos x }{ \sin x-\cos x } \times \cos x \]
\[\lim_{x\to \frac{\pi}{4}}\frac{\cancel { \sin x-\cos x }}{\cancel{ \sin x-\cos x }} \times \cos x \]
\[\lim_{x\to \frac{\pi}{4}} \cos x\]
Can you find the limit now?

Answer 3

@cali2 ???

Answer 4

I think so....0.999?

Answer 5

\[cos \frac{\pi} 4= \frac 1 {\sqrt 2}\]

Answer 6

Do you get it @cali2 ?

Answer 7

Oki, I got how you got the answer to cos pi/2. Can you break down your. I know tanx= sinx/cosx but I am not sure how you got to cancel

Answer 8

numerator and denominator are same that's why I cancelled
\[\lim_{x\to \frac{\pi}{4}}\frac{\cancel { \sin x-\cos x }}{\cancel{ \sin x-\cos x }} \times \cos x\]

Answer 9

and it's cos pi/4 not pi/2

Answer 10

That was a typo.
Okay, Thank you

Answer 11

Welcome, now you try the same things with other questions
Try to cancel the common terms