# Trigonometric Limit Question

• May 28th 2008, 11:18 PM
SportfreundeKeaneKent
Trigonometric 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 PM
Moo
Hello,

Quote:

Originally Posted by SportfreundeKeaneKent
Having problem with this one...tried out several different identities but (Headbang)

lim x-> -pi/2

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

Well, I'd transform it..

$\sec^2 x=\frac{1}{\cos^2 x} \implies \frac{1}{\sec^2 x}=\cos^2 x$

And $\tan^2 x=\frac{\sin^2 x}{\cos^2 x}$

------> $\frac{1+\tan^2 x}{\sec^2 x}=\left(1+\frac{\sin^2 x}{\cos^2 x}\right) \cdot \cos^2 x$

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 PM
squarerootof2
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 PM
Moo
Quote:

Originally Posted by squarerootof2
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.

Or you can just remember that we get the famous identity :

$\cos^2 x+\sin^2 x=1$

(Evilgrin)
• May 28th 2008, 11:50 PM
mr fantastic
Quote:

Originally Posted by squarerootof2
[snip]
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.

While it's true that the sum of these two limits is equal to 1, you might want to reconsider your stated values for each of the individual limits .......
• May 29th 2008, 06:45 AM
Soroban
Hello, SportfreundeKeaneKent!

Quote:

$\lim_{x\to\text{-}\frac{\pi}{2}}\frac{1 + \tan^2\!x}{\sec^2\!x}$
How about this identity: . $1 + \tan^2\!x \:=\:\sec^2\!x$

Then: . $\lim_{x\to\text{-}\frac{\pi}{2}}\frac{1 + \tan^2\!x}{\sec^2\!x} \;=\;\lim_{x\to\text{-}\frac{\pi}{2}} \frac{\sec^2\!x}{\sec^2\!x} \;=\;\lim_{x\to\text{-}\frac{\pi}{2}} (1) \;=\;1$