Articles of indeterminate forms

Evalutating $\lim_{x\to +\infty} \sqrt{x^2+4x+1} -x$

This question already has an answer here: Limits: How to evaluate $\lim\limits_{x\rightarrow \infty}\sqrt[n]{x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}}-x$ 5 answers

Is $0^\infty$ indeterminate?

Is a constant raised to the power of infinity indeterminate? I am just curious. Say, for instance, is $0^\infty$ indeterminate? Or is it only 1 raised to the infinity that is?

Find the value of : $\lim_{x \to \infty} \sqrt{4x^2 + 4} – (2x + 2)$

Possible Duplicate: Limits: How to evaluate $\lim\limits_{x\rightarrow \infty}\sqrt[n]{x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}}-x$ $$\lim_{x \to \infty} \sqrt{4x^2 + 4} – (2x + 2)$$ So, I have an intermediate form of $\infty – \infty$ and I tried multiplying by the conjugate; however, I seem to be left with another intermediate form of $\frac{\infty}{\infty}$ and wasn’t sure what else to to do. […]

Calculate $\lim_{x\to 0}\frac{1}{x^{\sin(x)}}$

Calculate $$\lim_{x\to 0}\dfrac{1}{x^{\sin(x)}}$$ I’m pretty much clueless here, only that there is L’hospital obviously here. Would appreciate any help.

Is $\frac00=\infty$? And what is $\frac10$? Are they same? Does it hold true for any constant $a$ in $\frac{a}0$

This question already has an answer here: Is it wrong to tell children that $1/0 =$ NaN is incorrect, and should be $∞$? 12 answers

Is it wrong to tell children that $1/0 =$ NaN is incorrect, and should be $∞$?

I was on the tube and overheard a dad questioning his kids about maths. The children were probably about 11 or 12 years old. After several more mundane questions he asked his daughter what $1/0$ evaluated to. She stated that it had no answer. He asked who told her that and she said her teacher. […]

Solve a seemingly simple limit $\lim_{n\to\infty}\left(\frac{n-2}n\right)^{n^2}$

$$\lim_{n\to\infty}\left(\frac{n-2}n\right)^\left(n^2\right)$$ Why does this go to 0? Why can I not just divide each item in the fraction by n and assume it would go to 1?

Why is $0^0$ also known as indeterminate?

This question already has an answer here: Zero to the zero power – is $0^0=1$? 24 answers

Why doesn't using the approximation $\sin x\approx x$ near $0$ work for computing this limit?

The limit is $$\lim_{x\to0}\left(\frac{1}{\sin^2x}-\frac{1}{x^2}\right)$$ which I’m aware can be rearranged to obtain the indeterminate $\dfrac{0}{0}$, but in an attempt to avoid L’Hopital’s rule (just for fun) I tried using the fact that $\sin x\approx x$ near $x=0$. However, the actual limit is $\dfrac{1}{3}$, not $0$. In this similar limit, the approximation reasoning works out.

Find $\lim_{n \to \infty} \left$ (a question asked at trivia)

My friend’s trivia league had this math question: $$\lim_{n \to \infty} \left[\frac{(n+1)^{n + 1}}{n^n} – \frac{n^{n}}{(n-1)^{n-1}} \right]$$ After computing a few values, one could guess the answer is $e$ = 2.718…But how can we prove that is the limit? Someone offered up a hand-wavy proof like this: \begin{align} \lim_{n \to \infty} \left[\frac{(n+1)^{n + 1}}{n^n} – […]