Help me on this

Note by Ayushadarsh Tiwari
1 year, 3 months ago

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  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

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\(\displaystyle \lim_{x\to\infty} \left(e^{3x}-x\right)^{\dfrac{1}{4x}}\) :

Solution: Whenever you have something raised to the power something and it looks pretty awkward, the first thing to do is to take logarithms. If you denote the limit by \(A\) then we have,

\(\begin{align} \log A &= \log\left(\lim_{x\to\infty} \left(e^{3x}-x\right)^{\dfrac{1}{4x}}\right) \\ &= \lim_{x\to\infty} \log\left(e^{3x}-x\right)^{\dfrac{1}{4x}} \\ &= \lim_{x\to\infty}\dfrac{1}{4x}\log(e^{3x}-x) \\ &= \lim_{x\to\infty} \dfrac{3(e^{3x}-1)}{4(e^{3x}-x)} \quad [\text{Using L-Hopital's Rule}] \\ &= \lim_{x\to\infty} \dfrac{3(1-e^{-3x})}{4(1-xe^{-3x})} \\ &= \dfrac{3}{4}\dfrac{1-\lim_{x\to\infty}e^{-3x}}{1-\lim_{x\to\infty}xe^{-3x}} \\ &=\dfrac{3}{4}\end{align}\)

The last two lines use that fact that \(\displaystyle \lim_{x\to\infty}e^{-3x}=0\) & \(\displaystyle \lim_{x\to\infty}\dfrac{x}{e^{3x}}=\lim_{x\to\infty}\dfrac{3}{e^{3x}}=0\) by L-Hopital's rule

The answer is hence \(\displaystyle \lim_{x\to\infty} \left(e^{3x}-x\right)^{\dfrac{1}{4x}}=e^{3/4}\)

Aditya Narayan Sharma - 1 year, 3 months ago

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