# Math Help - How to calculate this limit?

1. ## How to calculate this limit?

Hello forum members, thanks for all the help so far,
I would like to know how to calculate the limit below.

$\lim_{x\to1^+}{{55\,\sin \left(\pi\,x\right)}\over{1-x}}$

As you can see I've finally understood how to write equations using the math tool.

Thanks for the help.

2. Originally Posted by azarue
Hello forum members, thanks for all the help so far,
I would like to know how to calculate the limit below.

$\lim_{x\to1^+}{{55\,\sin \left(\pi\,x\right)}\over{1-x}}$

As you can see I've finally understood how to write equations using the math tool.

Thanks for the help.
i believe you have $\lim_{x\to1^+}{{55\,\sin \left(\pi\,x\right)}\over{1-x}}=0$ since the numerator is zero and the denominator is not.

3. ## I don't think so.

Because X ---> 1+

I can't calculate a limit with the form '0/0'.

4. Originally Posted by Raoh
i believe you have $\lim_{x\to1^+}{{55\,\sin \left(\pi\,x\right)}\over{1-x}}=0$ since the numerator is zero and the denominator is not.
$\lim_{x \to 1^+}1-x = 0$

OP: Do you know about L'Hopital's rule?

5. ## I know this rule, I just can't use it.

I went to an online limit calculator and it shows that :

$\lim_{x\to1^+}{{55\,\sin \left(\pi\,x\right)}\over{1-x}} = 55\,\pi$

How can I get this result without using L'L'Hopital's rule?

6. Originally Posted by azarue
Because X ---> 1+

I can't calculate a limit with the form '0/0'.
sorry what i wrote is wrong
put $h=1-x$ you'll find $\lim_{h\to 0}{{55\,\sin \left(\pi\,(1-h)\right)}\over{h}}=55\pi$
here's how
$\lim_{h\to 0}{{55\,\sin \left(\pi\,(1-h)\right)}\over{h}}$= $55 \lim_{ h \rightarrow 0}\frac{\sin(\pi-\pi h) }{ h}$= $55\lim_{ h \rightarrow 0}-\pi \frac{\sin(-\pi h) }{h \pi}=55\pi$