take limit without using trig identities

**yvonnehr** · October 16th 2012, 05:38 PM

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.

**skeeter** · October 16th 2012, 05:45 PM

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?

**yvonnehr** · October 16th 2012, 06:31 PM

I see. I didn't understand that that's what the hint meant. Thank you. Ill try that.

**yvonnehr** · October 16th 2012, 10:17 PM

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.

**Vlasev** · October 16th 2012, 11:42 PM

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 $x \to 0$ ?

**skeeter** · October 17th 2012, 03:13 AM

no limit "manipulation" or identities are required here, only recognition of the definition of a derivative ...

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

$\lim_{h \to 0} \frac{\tan(y+h) - \color{red}{\tan{y}}}{h} = \sec^2{y}$

**hollywood** · October 17th 2012, 07:09 AM

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

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