Thread: Evaluating Limits Analyticall

1. Evaluating Limits Analyticall

Determine the limit of the trigonometric function (if it exists).

1. lim sin x / 5x
(x -> 0)
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$lim sin x / 5x
(x -> 0)

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(x -> 0)
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2. lim tan^2x / x
(x ->0)

3. lim cos x tan x / x
(x -> 0)

Below, find lim f(x+h) - f(x) / h
----------- (x->0)

1. f(x) = radical x

THANK YOU

2. Originally Posted by Amadeus
no one? T_T
be patient
Originally Posted by Amadeus
Determine the limit of the trigonometric function (if it exists).

1. lim sin x / 5x
(x -> 0)
$lim sin x / 5x
(x -> 0)$
Note that: $\lim_{x \to 0} \frac {\sin x}{5x} = \frac 15 \lim_{x \to 0} \frac {\sin x}x$

you should know what $\lim_{x \to 0} \frac {\sin x}x$ is, so continue.

2. lim tan^2x / x
(x ->0)
multiply the top and bottom by x:

we get: $\lim_{x \to 0} \frac {x \tan^2 x}{x^2} = \lim_{x \to 0} x \left( \frac {\tan x}{x} \right)^2$

Now, $\lim_{x \to 0} \frac {\tan x}x = \lim_{x \to 0} \frac {\sin x}x$, so you should be able to finish this one too

3. lim cos x tan x / x
(x -> 0)
$\lim_{x \to 0} \cos x \left( \frac {\tan x}x\right) = \lim_{x \to 0}\cos x \cdot \lim_{x \to 0} \frac {\tan x}x$

same story as the last one. perhaps a nicer way to do it is to realize that $\tan x = \frac {\sin x}{\cos x}$, so the limit would simplify to: $\lim_{x \to 0} \frac {\sin x}x$

...i guess you're starting to see how important the limit $\lim_{x \to 0} \frac {\sin x}x$ is, huh? ...you may want to simplify this way in the last problem

Below, find lim f(x+h) - f(x) / h
----------- (x->0)

1. f(x) = radical x
You want: $\lim_{h \to 0} \frac {\sqrt{x + h} - \sqrt {x}}h$

rationalize the numerator and simplify

3. ^ Thank you so much, Jhevon! It really helped me a lot.

But I'd like to check to see if my answers are right.

1. $1/5$
2. $1$
3. $0$ <-- I think that one is wrong...

--

1. 1 / radical h

4. Originally Posted by Amadeus
^ Thank you so much, Jhevon! It really helped me a lot.

But I'd like to check to see if my answers are right.

1. $1/5$
yes

2. $1$
No, for small $x\ \tan(x) \approx x$ so $(\tan(x))^2/x \approx x$, so the limit is $0$

3. $0$ <-- I think that one is wrong...
Again for small $x\ \cos(x) \approx 1\ \tan(x) \approx x$ so $[\cos(x) \tan(x)]/x \approx 1$
the limit is $1$

RonL

5. Originally Posted by Amadeus
^ Thank you so much, Jhevon! It really helped me a lot.

But I'd like to check to see if my answers are right.

1. $1/5$
CaptainBlack already told you this was correct

2. $1$
remember, we were multiplying by an x here, and the x went to zero, so the second limit went to 1 and we multiplied by zero

3. $0$ <-- I think that one is wrong...
here, you seem to have caught the fact that the x made us multiply by zero, however, the x was a typo (sorry) it should have been cos(x). (you should have caught that though, and i also mentioned that the limit simplified to sin(x)/x, so that's a second thing you should have caught)

1. 1 / radical h
you should get $\frac 1{2 \sqrt {x}}$ (h is going to zero, there should be no h in your final answer, did you try my suggestion?)

6. Thank you very much! I understand (for the most part)

Although I got stuck on this one...

Below, find lim f(x+h) - f(x) / h
----------- (x->0)

$f(x) = 4/x$

I tried it and got the first step as:

(4/x+h) - (4/x) / h

At this step, I'm pretty sure I'm on the wrong track but I multiplied both top & bottom by (4/x + h) + (4/x)

Again, it turns out weird and I'm lost...

7. Originally Posted by Amadeus
Thank you very much! I understand (for the most part)

Although I got stuck on this one...

Below, find lim f(x+h) - f(x) / h
----------- (x->0)

$f(x) = 4/x$

I tried it and got the first step as:

(4/x+h) - (4/x) / h

At this step, I'm pretty sure I'm on the wrong track but I multiplied both top & bottom by (4/x + h) + (4/x)

Again, it turns out weird and I'm lost...
Your limit has a typo. It should be as h goes to 0.

This one simply has a lot of algebra involved:
$\lim_{h \to 0}\frac{\frac{4}{x + h} - \frac{4}{x}}{h}$

Subtract the fractions:
$= \lim_{h \to 0}\frac{\frac{4x}{x(x + h)} - \frac{4(x + h)}{x(x + h)}}{h}$

$= \lim_{h \to 0}\frac{4x - 4(x +h)}{hx(x + h)}$

$= \lim_{h \to 0}\frac{-4h}{hx(x + h)}$

$= \lim_{h \to 0}\frac{-4}{x(x + h)}$

Now take the limit:
$= \frac{-4}{x^2}$

-Dan