Find the Limit...its TRIG!!!!! Please Help!!!!!! ASAP

[DancingQueen09]

I need some help can you please show steps and explain in detail how to do everything... thanks!!!!

Lim Tan(3x) / Tan(4x)
x->0

(tan3x over tan4x)

and also this question PLEASE

Lim

X->10

Thanks everyone in advance

 

[mmm4444bot]

On the first exercise, it seems intuitive that the limit is 3/4.

Are you familiar with the following approximations?

sin(x) ? x, when x is near zero

cos(x) ? 1, when x is near zero.

If you factor tan(3x) and tan(4x) into ratios of sines and cosine, and then take a numerical approach by letting x get close to zero in steps 0.01, 0.001, 0.001, 0.0001, you'll see that the given expression is approaching 3/4. At least, it did in my head just now. You'd better check that, by writing something down.

On the second exercise, you can evaluate by substitution; it's a well-behaved function around x = 10, yes?

 

[BigGlenntheHeavy]

\(\displaystyle \lim_{x\to0} \ \frac{tan(3x)}{tan(4x)} \ = \ \lim_{x\to0} \ \frac{3sec^{2}(3x)}{4sec^{2}(4x)} \ = \ \frac{3}{4}.\)

\(\displaystyle The \ Marqui \ came \ into \ play.\)

 

[galactus]

I will try the non-L'Hopital method then.

\(\displaystyle \frac{tan(3x)}{tan(4x)}=\frac{cos(6x)+cos(4x)+cos(2x)}{cos(6x)+cos(4x)+cos(2x)+1}\)

Now, simply plug in x=0 and we get \(\displaystyle \frac{3}{4}\)

 

[soroban]

Hello, DancingQueen09!

Here's another non-L'Hopital approach.
\(\displaystyle \text{I }hammered\text{ it into the form: }\:\lim_{\theta\to0}\frac{\sin\theta}{\theta} \:=\:1\)

\(\displaystyle \lim_{x\to0}\frac{\tan(3x)}{\tan(4x)}\)

\(\displaystyle \text{We have: }\:\frac{\dfrac{\sin(3x)}{\cos(3x)}}{\dfrac{\sin(4x)}{\cos(4x)}} \;=\; \frac{\sin(3x)}{\sin(4x)}\cdot\frac{\cos(4x)}{\cos(3x)}\)


\(\displaystyle \text{Multiply by: }\,\frac{12x}{12x} \,=\,\frac{3}{4}\!\cdot\!\frac{4x}{3x}\qquad\frac{3}{4}\cdot\frac{4x}{3x}\cdot\frac{\sin(3x)}{\sin(4x)}\cdot\frac{\cos(4x)}{\cos(3x)}\)


\(\displaystyle \text{And we have: }\;\frac{3}{4}\cdot\frac{\sin(3x)}{3x}\cdot\frac{4x}{\sin(4x)}\cdot\frac{\cos(4x)}{\cos(3x)}\)


\(\displaystyle \text{Take the limit: }\;\lim_{x\to0}\left[\frac{3}{4}\cdot\frac{\sin(3x)}{3x}\cdot\frac{4x}{\sin(4x)}\cdot\frac{\cos(4x)}{\cos(3x)}\right]\)

. . . . . . . . . \(\displaystyle =\;\frac{3}{4}\cdot\underbrace{\left[\lim_{x\to0}\frac{\sin(3x)}{3x}\right]}_{1} \cdot\underbrace{\left[\lim_{x\to0}\frac{4x}{\sin(4x)}\right]}_{1}\cdot\underbrace{\left[\lim_{x\tp0}\frac{\cos(4x)}{\cos(3x)}\right]}_{\frac{1}{1}} \;=\;\frac{3}{4}\)