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CC 1355 RULE to SHOW CAUSE PAGE 1 USING THIS FORM 1
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FAQs
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Can I fill out the NIOS April 2018 stream 1 form by an offline form because the online form is closed and I want to appear on this stream to save this question?
You can contact your respective regional centers for this. Though on the website there is no where mentioned that the application form will be available offline. However, it is best to ask them regarding this. -
How would you use LHopital's rule to show that the [math]\lim_{x \to 0} \frac{(1 + \frac{x}{2}) \log (x + 1) - x}{x^3} = \frac{1}{12}[/math]?
L'Hôpital's rule says that when we seek the limit of a ratio [math]f(x)/g(x)[/math], and substitution gives an indeterminate result of 0/0 (or [math]\infty/\infty[/math]), we can instead seek the limit of [math]f'(x)/g'(x)[/math]. Furthermore, if the limit is still indeterminate after applying that rule, we can do so repeatedly.Because this looks like a homework question, I’ll show the process but leave out the details of finding the derivatives and the algebra required to rewrite the expressions:[math]\begin{align*}&\phantom{={}}\lim_{x \to 0} \frac{(1 + \frac{x}{2}) \ln (x + 1) - x}{x^3}& \text{indeterminate (0/0)}\\ &=\lim_{x \to 0} \frac{(2 + x) \ln (x + 1) - 2x}{2x^3}\\ &=\lim_{x \to 0} \frac{(2+x)\frac{1}{x+1}+ \ln (x + 1) - 2}{6x^2} & \text{L'Hôpital #1}\\ &=\lim_{x \to 0} \frac{(x+1)\ln (x + 1) - x}{6x^2(x+1)} & \text{(still indeterminate)}\\ &=\lim_{x \to 0} \frac{\ln (x + 1)}{18x^2+12x} & \text{L'Hôpital #2}\\ &=\lim_{x \to 0} \frac{\frac{1}{x + 1}}{36x+12} & \text{L'Hôpital #3}\end{align*}\tag*{}[/math]After the third application of L'Hôpital’s rule, the resulting expression is no longer indeterminate, so the limit can be evaluated by direct substitution.How would you use LHopital's rule to show that the [math]\lim_{x \to 0} \frac{(1 + \frac{x}{2}) \log (x + 1) - x}{x^3} = \frac{1}{12}[/math]? -
Is every prime number other than 2 and 3 of the form (6k±1)? Is this a proven result? What are other resources about it?
Every integer can be written in the form of [math]6n,6n+1,6n+2,6n+3,6n+4[/math] or [math]6n+5[/math].It is quite evident that [math]6n[/math] is always divisible by 6, 6n+1 may be prime, [math]6n+2[/math] will always be divisible by 2, [math]6n+3[/math] by 3, and [math]6n+4[/math] by 2. [math]6n+5[/math] may be prime. Our candidates for primality are therefore [math]6n+1[/math] and [math]6n+5[/math]. We can then see that [math]6n+5=6(n+1)-1=6x-1[/math] Therefore, every prime can be written in the form of [math]6n \pm 1[/math] -
I’ve been staying out of India for 2 years. I have an NRI/NRO account in India and my form showed TDS deduction of Rs. 1 lakh. Which form should I fill out to claim that?
The nature of your income on which TDS has been deducted will decide the type of ITR to be furnished by you for claiming refund of excess TDS. ITR for FY 2017–18 only can be filed now with a penalty of Rs. 5000/- till 31.12.2018 and Rs. 10,000/- from 01.01.2019 to 31.03.2019. So if your TDS relates to any previous year, then just forget the refund. -
What is [math]\lim_{w \to -\infty} \frac{w^2 - 4w + 1}{3w^2 + 7w - 4}[/math]? This is a problem about indeterminate forms using L’Hôpital’s rule.
When calculating limits of dividing polynomials approaching infinity (either one), there’s a rule you can use. The Polynomial Growth rule. The monomial which bears the highest order will govern the entire curve’s growth, so you can neglect the rest and work with the highest-powered monomial.When you apply that rule to this particular limit, you’ll get:[math]\lim_{x \to \infty} = \frac{w^2}{3w^2}[/math]Which will give you 1/3. You don’t need L’Hopital’s rule to solve it. If using it is a requirement, though, Bruce Walden’s answer is the absolute way to go. Hope it helps.EDIT: Thanks to Robby Goetschalckx for making the equations look as they’re supposed to look, hahahaha. I haven’t mastered Quora’s math writing. If anyone is willing to offer a crash course I’d be very thankful. -
How do I evaluate this limit [math]\displaystyle \lim_{x\to0}\left(\frac{1}{\log(x+\sqrt{1+x^2})}-\frac{1}{\log(1+x)}\right)[/math]? You are free to use L'Hôpital's rule.
One could do it with power series a bit easier than with L’Hospial’s rule.The point is that to second order in x the denominators are log(1+x) = x -x^2/2+… and log(1+x+x^2/2 + …) = x+x^2/2 -1/2(x+x^2/2)^2 +… = x+…So you get 1/(x+0x^2+…) - 1/(x-x^2/2+…) = (-x^2/2+…)/(x^2+…) which approaches -1/2.I am skipping details and justification, but hopefully you can fill them in.
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