# Thread: limits of a vector function

1. ## limits of a vector function

Hello, the question is:

limit t-->0+ (cos(t), sin(t), tln(t))

(the limit as t approaches 0 from the right)

Thanks,
-cb

2. Originally Posted by sprinks13
Hello, the question is:

limit t-->0+ (cos(t), sin(t), tln(t))

(the limit as t approaches 0 from the right)

Thanks,
-cb
Just pass the limit on each term.

3. yup, I got (1,0,?) i can't figure out the tln(t) part.

Doesn't ln(t) approach infinity when t approaches 0? And then it is 0+ multiplied by infinity?

4. Originally Posted by sprinks13
yup, I got (1,0,?) i can't figure out the tln(t) part.

Doesn't ln(t) approach infinity when t approaches 0? And then it is 0+ multiplied by infinity?
note that $t \ln t = \frac {\ln t}{\frac 1t}$. use L'Hopital's rule (i know everyone hates me for saying that, but i don't care) (and lnt goes to -infinity as t goes to 0)

5. Originally Posted by Jhevon
(i know everyone hates me for saying that, but i don't care)
I am working on one approach without using L'Hopital.
(This is not for the original poster. Do not confuse yourself. This is just for fun for other posters).
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Directly by definition.
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We want to show,
$\lim_{t\to 0^+} t\ln t = \lim_{t\to 0^+} t\int_1^t \frac{1}{x} dx = \lim_{t\to 0^+} - t\int_t^1 \frac{1}{x} dx$

Ignore the minus sign for now. Now try to show that $\lim_{t\to 0^+} t\int_t^1 \frac{1}{x} \ dx = 0$.

Let $t$ be small enough so that we are working on the interval $[t,1]$.

Now, $\frac{1}{x} \leq \frac{1}{e^x}$ on $[t,1]$ because $e^x \geq x$ on $(0,1]$. Thus, $\int_t^1 \frac{1}{x} dx \leq \int_1^t \frac{1}{e^x} dx = \frac{1}{e^t}-1$.

This means,
$\left| t\int_t^1 \frac{1}{x} dx \right| \leq \frac{t}{e^t} - t \leq \frac{t}{e^t}$.
But,
$\lim_{t\to 0^+} \frac{t}{e^t} = 0$.

No L'Hopital!