Question

\[1) \lim_{x \rightarrow 0} \frac{ \sin(\tan(5x)) }{ 4x }\]
\[2) \lim_{x \rightarrow 0} \frac{ \tan(\sin(7x))}{ 3x }\]
\[3) \lim_{x \rightarrow 0} \sin(7x(3+\frac{ 2 }{ x }))\]

• Write the step, Thanks.

Answer 1

\(1. \lim\limits_{x\to0}: \cfrac{\sin(\tan(5x))}{4x} \left[\frac00\right] = \cfrac{\cos(\tan(5x))\cdot\sec^2(5x)\cdot5}{4} = \cfrac{1\cdot1\cdot5}{4} = \cfrac54\)
The result could alternatively be _motivated_ as folows: Near x=0, both sin(x) and tan(x) are roughly x, so sin(tan(5x)) is roughly 5x, and 5x/4x = 5/4.

Answer 2

The \(\left[\frac00\right]\) means that we will use l'Hôpital's rule.

Answer 3

Yeah, but sin(tan(5x)) -> cos(tan(5x))⋅sec2(5x)⋅5 What did you do to get that ?

Answer 4

I used the chain rule.

Answer 5

That's it thanks :D.

Answer 6

No problem!

Answer 7

Do you know the answer of the last one ?

Answer 8

The second is almost the same as the first.

Answer 9

Does \(\sin7x\,(3+\frac2x)\) mean \((\sin7x)(3+\frac2x)\) or \(\sin(7x\,(3+\frac2x))\)?

Answer 10

The second one

Answer 11

sin(7x(3+2x))

Answer 12

\[\lim\limits_{x\to0}: \sin(7x\,(3+\frac2x)) = \sin(21x+14) = \sin(14)\]

Answer 13

I found that it equals to sin(14) but isn't it strange i really never got something like that in limits

Answer 14

It happens sometimes. A month ago wolframalpha evaluated this for me: \(\lim\limits_{x\to\infty} \left(\cfrac{(x+i)^2}{x^2+1}\right)^x = e^{2i}\).

Answer 15

A nice problem , indeed.

Answer 16

Wow, I just realized how to evaluate it! I'll show you :)

Answer 17

Lol , Is it too long like that ? 0_0

Answer 18

I'm using one definition of \(e^x\) saying that \(e^x = \lim\limits_{n\to\infty} (1+x/n)^n\).
\[\begin{split}
\lim\limits_{x\to\infty} \left(\cfrac{(x+i)^2}{x^2+1}\right)^x
&= \lim\limits_{x\to\infty} \left(\cfrac{(x+i)^2/{x^2}}{(x^2+1)/{x^2}}\right)^x\\
&= \lim\limits_{x\to\infty} \left(\cfrac{(1+i/x)^2}{1+1/{x^2}}\right)^x\\
&= \lim\limits_{x\to\infty} \frac{(1+i/x)^{2x}}{(1+1/{x^2})^x}\\
&= \frac{\lim\limits_{x\to\infty}(1+i/x)^{2x}}{\lim\limits_{x\to\infty}(1+1/{x^2})^x}\\
\end{split}\]
The numerator is
\(\qquad\lim\limits_{x\to\infty}((1+i/x)^x)^2 = (e^x)^2 = e^{2x}\)
and the denominator is
\(\qquad\lim\limits_{x\to\infty}{(1+(1/x)/x})^x = \lim\limits_{x\to\infty} e^{1/x} = 1\) (This part could be made more formal.)
so the quotient is \(e^{2x}\).

Answer 19

limx→∞e1/x=1 (This part could be made more formal.)
so the quotient is e2x.

I can't understand this part the rest is fine to me.

Answer 20

Sorry the \((e^x)^2\) etc should have an \(i\) instead of \(x\).

Answer 21

I'll try to do it formally.

Answer 22

"Sorry the (ex)2 etc should have an i instead of x."
I didn't notice this mistake too because of the x in the saying.

What's the hardest limit problem ? I wanna test myself.

Answer 23

Well you could try to evaluate that denominator because I can't seem to figure it out.

Answer 24

OII I will try.

Answer 25

Any clues mathmate?

Answer 26

Still working on it.

Answer 27

I'll check back later.

Answer 28

Okie.

Answer 29

Here is how my teacher did it:
\[\qquad\begin{split}
L &= \lim\limits_{x\to\infty}{(1+1/x^2})^x \\
&= \lim\limits_{x\to\infty}{e^{\ln\left((1+1/x^2)^x\right)}} \\
&= \lim\limits_{x\to\infty}{e^{x\ln(1+1/x^2)}} \\
&= e^y\; (\text{where } y = \lim\limits_{x\to\infty}{x\ln(1+1/x^2)}) \\
&\qquad\begin{split}
y &= \lim\limits_{x\to\infty} \frac{\ln(1+1/x^2)}{1/x} {\small\left[\frac00\right]}\\
&= \lim\limits_{x\to\infty} \frac{\left(\frac{-2/x^3}{1+1/x^2}\right)}{-1/x^2}\\
&= \lim\limits_{x\to\infty} \frac{2}{x+1/x}\\
&= 0
\end{split}\\
L &= e^0 = 1
\end{split}\]