Having problem with this one...tried out several different identities but
lim x-> -pi/2
1 + (tan x)^2/ (sec x)^2
Hello,
Well, I'd transform it..
$\displaystyle \sec^2 x=\frac{1}{\cos^2 x} \implies \frac{1}{\sec^2 x}=\cos^2 x$
And $\displaystyle \tan^2 x=\frac{\sin^2 x}{\cos^2 x}$
------> $\displaystyle \frac{1+\tan^2 x}{\sec^2 x}=\left(1+\frac{\sin^2 x}{\cos^2 x}\right) \cdot \cos^2 x$
Continue ?
(I don't know if 1+ is in the fraction or not... But the trick would be the same)
observe that [1+(tanx)^2]/(secx)^2=1/(secx)^2+(tanx)^2/(secx)^2
this is equal to 1/(secx)^2 + [(sinx)^2/(cosx)^2]/[1/(cosx)^2]. (this is because (tanx)^2=(sinx)^2/(cosx)^2 and (secx)^2=1/(cosx)^2)
here dividing by 1/(cosx)^2 is same as multiplying by (cosx)^2, which cancels out the denominator of first term.
Hence we have that your original function is equal to [1/(secx)^2]+(sinx)^2. taking the limit of that as it approaches -pi/2, we can literally "plug in" -pi/2 into this function to see that 1/(secx)^2 = 1/2 at x=-pi/2 and (sinx)^2 = 1/2 at x=-pi/2. thus, adding it up we get that limit approachs 1.
Hello, SportfreundeKeaneKent!
How about this identity: .$\displaystyle 1 + \tan^2\!x \:=\:\sec^2\!x$$\displaystyle \lim_{x\to\text{-}\frac{\pi}{2}}\frac{1 + \tan^2\!x}{\sec^2\!x}$
Then: .$\displaystyle \lim_{x\to\text{-}\frac{\pi}{2}}\frac{1 + \tan^2\!x}{\sec^2\!x} \;=\;\lim_{x\to\text{-}\frac{\pi}{2}} \frac{\sec^2\!x}{\sec^2\!x} \;=\;\lim_{x\to\text{-}\frac{\pi}{2}} (1) \;=\;1 $