Having problem with this one...tried out several different identities but (Headbang)

lim x-> -pi/2

1 + (tan x)^2/ (sec x)^2

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- May 28th 2008, 11:18 PMSportfreundeKeaneKentTrigonometric Limit Question
Having problem with this one...tried out several different identities but (Headbang)

lim x-> -pi/2

1 + (tan x)^2/ (sec x)^2 - May 28th 2008, 11:26 PMMoo
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) - May 28th 2008, 11:32 PMsquarerootof2
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. - May 28th 2008, 11:40 PMMoo
- May 28th 2008, 11:50 PMmr fantastic
- May 29th 2008, 06:45 AMSoroban
Hello, SportfreundeKeaneKent!

Quote:

$\displaystyle \lim_{x\to\text{-}\frac{\pi}{2}}\frac{1 + \tan^2\!x}{\sec^2\!x}$

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 $