Question

Find the limit as x approaches pi/2 for ((sinx/cos(^2)x)-tan(^2)x)

Answer 1

\[\Large\rm \lim_{x\to \pi/2}\frac{\sin x}{\cos^2x}-\tan^2 x\]

Mmmm thinking.

Answer 2

L'Hospital's rule is okay to use?

Answer 3

\[
\large \lim_{x\to \pi/2}\frac{\sin x}{\cos^2x}-\tan^2 x =
\large \lim_{x\to \pi/2}\frac{\sin x}{\cos^2x}-\frac{\sin^2 x}{\cos^2 x} = \\
\large \lim_{x\to \pi/2}\frac{\sin x(1-\sin x)}{\cos^2x} = ?
\]

Answer 4

0/0 form. Try L'H a couple of times.

Answer 5

unfortunately we cannot use l'hospital rule

Answer 6

is there any other way to solve it? this one has me stumped!

Answer 7

To finish where @aum stopped...
\[\lim_{x\rightarrow\pi/2}\dfrac{\sin x(1-\sin x)}{\cos^2x}\\~\\\lim_{x\rightarrow\pi/2}\dfrac{\sin x(1-\sin x)}{1-\sin^2x}\\~\\\lim_{x\rightarrow\pi/2}\dfrac{\sin x(1-\sin x)}{(1+\sin x)(1-\sin x)}\\~\\\lim_{x\rightarrow\pi/2}\dfrac{\sin x}{1+\sin x}\]

Answer 8

From here, you can make direct substitution.

Answer 9

Is geerky42's solution using L'hopital's rule, or is it without?

Answer 10

Without.

Answer 11

Without. Its just clever factoring.

Answer 12

ok thank you! for your help!

Answer 13

I replaced \(\cos^2x \) to \(1-\sin^2 x\) because \(\sin^2 x + \cos^2 x = 1\)

Answer 14

Welcome

Answer 15

what did you replace for tan^2x

Answer 16

\(\dfrac{\sin^2 x}{\cos^2 x}\)

Answer 17

What happened to the sin^2x from the tan^2x?

Answer 18

\[\lim_{x\rightarrow\pi/2}\dfrac{\sin x}{\cos^2x} - \dfrac{\sin^2x}{\cos^2x} = \lim_{x\rightarrow\pi/2}\dfrac{\sin x-\sin^2 x}{\cos^2x}\]
Then we factored out \(\sin x\) in numerator to get this: \[\lim_{x\rightarrow\pi/2}\dfrac{\sin x(1-\sin x)}{\cos^2x}\]
Does that help?

Answer 19

Yes it does thank you, again!

Answer 20

No problem