# L'Hopital's Lim x->0 tan x / x

I've done this problem but am not comfortable with the answer. Could someone take a quick look?

Limx->0 tan x /x

Limx->0 sec^2x ==> 0

I'm just not sure this is right...

quasar987
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sec(x)=1/cos(x)

So how do you conclude that the limit of sec²(x) is 0 ?

ok, so could I use sec^2(x) = 1/cos^2(x) then use the power reducing formula to make: limx->0 2/(1+cos2x)? Then as x->0, then the value would be 2/1 thus 2?

quasar987
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Do you think that cos(0)=0 ? That's sin's job.

ahhh cos(0) = 1 and sin(0) = 0. I should have verified that b4... Given this, then 2/(1+cos2(0)) = 2/2 = 1. Knowing this, there are 2 things that I should have worked through: secx = 1/cosx and that cos(0) = 1.

Thanks very much for your help!

mathwonk
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or just recall that sin/cos = tan and cos(0)= 1, so tan(x)/x has the same limit as sin(x)/x which everyone knows is 1.

but still everything relies on knowing cos(0) =1.

Thank you! I'm not quite sure I understand, but I get tan = sin/cos since tye hypenuses cancel but how can I prove tan(x)/x has the same limit as sin(x)/x? I need to read and understand some of the finer points, or I won't be able to answer some fundamental questions. Any suggestions as to where to start? I'm going to google it and try and find something. Having been away from this stuff for a very long time, it's tough to come back....

quasar987
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The keys to mathwonk's argument is that

$$\frac{\tan(x)}{x}=\frac{1}{\cos(x)}\frac{\sin(x)}{x}$$

, that

$$\lim_{x\rightarrow 0}\frac{\sin(x)}{x}=1$$

, and that

$$\lim_{x\rightarrow a}f(x)g(x)=\lim_{x\rightarrow a}f(x)\cdot \lim_{x\rightarrow a}g(x)$$

(provided the last two limits exist).

Gib Z
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Have you learned Taylor Series yet? Since this is the limit as x goes to zero, the first term of tan x's taylor series Could have been substituted in and The limit would have been easy to see.

I understand the first part since (opp/hyp)/(adj/hyp) thus opp/adj = tan. That part makes sense. I don't get sin(x)/x = 1. Is this just something I should accept? I understand sin^2+cos^2=1. Would that be part of it? As for the last part, that makes sense provided they both approach the limit from the same side.

I found some articles on the web that sort of explain things, but I suspect that I'm going to have to read them a few times to really understand it!

Have you learned Taylor Series yet? Since this is the limit as x goes to zero, the first term of tan x's taylor series Could have been substituted in and The limit would have been easy to see.

No, we have not reviewed Taylor Series yet. We have just started that section. Might there be an area in which I could review and get a jump on what to expect?

THANKS!

Gib Z
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I understand the first part since (opp/hyp)/(adj/hyp) thus opp/adj = tan. That part makes sense.[\QUOTE]

No it doesn't. I know Im being pedantic, but you really need an argument. tan of what? 5 bananas?

I don't get sin(x)/x = 1. Is this just something I should accept?
[\QUOTE] NO. Because thats not true for all values of x. What they were talking about is $\lim_{x\to 0} \frac{\sin x}{x} =1$
I understand sin^2+cos^2=1. [\QUOTE]
Good

Just search Taylor Series In Google and Wikipedia.

Gib Z
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Quotes thing doesn't seem to want to work for me, But you get the point.

I do get ur point: tan x = opp/adj As for point 2: limx->0 sin(x)/x = 1 vs. sin(x)/x = 1 ARE 2 different statements. Good point and I need to keep this in mind since they say very different things.

I'll search now in google for 'Taylor Series'. IK Wikipedia has also been a source of good info for me too!

Thanks!

quasar987