# take limit without using trig identities

• October 16th 2012, 05:38 PM
yvonnehr
take limit without using trig identities
Will someone help me solve this? I tried to solve many times and keep getting 0 as the answer. Wolfram Alpha says the answer should be (secy)^2.

Without using any trigonometric identities, find

lim x->0 [tan(x + y) - tan y]/x

Hint: Relate the given limit to the definition of the derivative of an appropriate function of y.

I used tan(x + y) = (tan x + tan y)/(1 - tan x tan y). I still got 0 as the answer. (Thinking)
• October 16th 2012, 05:45 PM
skeeter
Re: take limit without using trig identities
use the hint ... change x to h and use the def. of a derivative

$\lim_{h \to 0} \frac{\tan(y+h) - \tan{y}}{h}$

look familiar?
• October 16th 2012, 06:31 PM
yvonnehr
Re: take limit without using trig identities
I see. I didn't understand that that's what the hint meant. Thank you. Ill try that.
• October 16th 2012, 10:17 PM
yvonnehr
Re: take limit without using trig identities
Initially I used the definition without changing the variables, i.e. So in a way, I had used the instruction which stated to use the definition of a derivative... I'm not sure that I used the hint in the way you referred.

$\lim_{x \to \0} \frac{(tan x + y) - tan y}{x}$

$=\lim_{x \to \0} \frac{\frac{tan x + tan y}{1 - tan x tan y} - tan y}{x}$

$=\lim_{x \to \0} \frac{tan x + tan y - tan y(1 - tan x tan y)}{1 - tan x tan y}\div x$

$=\lim_{x \to \0} \frac{tan x + tan y - tan y + tan x tan^2 y}{x(1 - tan x tan y)}$

$=\lim_{x \to \0} \frac{tan x (1 + tan^2 y)}{x(1 - tan x tan y)}$

At this point, I don't know what to do. I can't use L' Hopital's Rule yet because that is introduced later in the book.
... some middle steps

$=sec^2y$ per Wolfram Alpha, not the solution's manual.
• October 16th 2012, 11:42 PM
Vlasev
Re: take limit without using trig identities
Notice that

$\lim_{x\to 0} \frac{\tan x}{x} = \lim_{x\to 0} \frac{sin x}{x \cos x} = 1$

Since $\cos 0 = 1$ and the limit of $\sin(x)/x$ is 1. This means that your limit is equal to

$\lim_{x\to 0} \frac{1+\tan^2y}{1-\tan x \tan y}$

What happens to this expression as $x \to 0$?
• October 17th 2012, 03:13 AM
skeeter
Re: take limit without using trig identities
no limit "manipulation" or identities are required here, only recognition of the definition of a derivative ...

$\lim_{h \to 0} \frac{f(y+h) - \color{red}{f(y)}}{h} = f'(y)$

$\lim_{h \to 0} \frac{\tan(y+h) - \color{red}{\tan{y}}}{h} = \sec^2{y}$
• October 17th 2012, 07:09 AM
hollywood
Re: take limit without using trig identities
You said no trig identities, so the procedure you and Vlasev suggest, though valid, was not allowed.

You end up with the same result as Vlasev if you recognize

$\lim_{x \to \0} \frac{tan x (1 + tan^2 y)}{x(1 - tan x tan y)}$

$=\lim_{x \to \0} \frac{tan x}{x}\frac{1 + tan^2 y}{1 - tan x tan y}$

$=\lim_{x \to \0} \frac{tan x}{x} \lim_{x \to \0}\frac{1 + tan^2 y}{1 - tan x tan y}$

and the first factor is 1. But you only know that because of L'Hôpital's rule. You should see how to get from here to $\sec^2x$.

But the answer to the problem is given by skeeter - twice. And he actually uses the hint, which is always a good sign that you're on the right track. Do you see how skeeter's solution works?

- Hollywood