AShortcuts Tech
It's not working completely. Sometimes it does, but sometimes it does not. I think it worked better in v0.24 but am not certain
outputted tex:
Certainly! Let's consider the limit:
\[
\lim_{{x \to \infty}} (\sqrt{x^2 + x} - x)
\]
At first glance, if you substitute \(x \to \infty\), both \(\sqrt{x^2 + x}\) and \(x\) approach infinity, leading to the indeterminate form ∞ - ∞. To resolve this, we can use algebraic manipulation:
  1. Multiply and Divide by Conjugate:
\[
\sqrt{x^2 + x} - x = \frac{(\sqrt{x^2 + x} - x)(\sqrt{x^2 + x} + x)}{\sqrt{x^2 + x} + x}
\]
  1. Simplify the Numerator:
\[
(\sqrt{x^2 + x} - x)(\sqrt{x^2 + x} + x) = (\sqrt{x^2 + x})^2 - x^2 = x^2 + x - x^2 = x
\]
  1. Rewrite the Expression:
\[
\lim_{{x \to \infty}} \frac{x}{\sqrt{x^2 + x} + x}
\]
  1. Divide by \(x\):
\[
\lim_{{x \to \infty}} \frac{1}{\sqrt{1 + \frac{1}{x}} + 1}
\]
  1. Evaluate the Limit:
\(\frac{1}{x} \to 0\) as \(x \to \infty\), so:
\[
\lim_{{x \to \infty}} \frac{1}{\sqrt{1 + 0} + 1} = \frac{1}{1 + 1} = \frac{1}{2}
\]
Thus, the limit is \(\frac{1}{2}\). This demonstrates how the indeterminate form ∞ - ∞ can be resolved through clever algebraic manipulation to yield a determinate result.