If you find any mistakes, please make a comment! Thank you.

## Compute limits using L’Hospital’s Rule II

Solution:

### Part a

We can apply L’Hospital’s Rule repeatedly to get the answer. Please check the conditions for applying L’Hospital’s Rule carefully for every step. We have
\begin{align*}
\lim_{x\to 0}\frac{x^3}{\sin x-x}
=&\ \lim_{x\to 0}\frac{3x^2}{\cos x-1}\\
=&\ \lim_{x\to 0}\frac{6x}{-\sin x}\\
=&\ \lim_{x\to 0}\frac{6}{-\cos x}\\
=&\ \frac{6}{-1}=-6.
\end{align*}

### Part b

We shall use the following limit from Example 1, that is
$$\lim_{x\to 0}\frac{\sin x}{x}=1.$$ This also implies that
\label{eq:30-2}
\lim_{x\to 0}\frac{\tan x}{x}=\lim_{x\to 0}\frac{\sin x}{x}\lim_{x\to 0}\frac{1}{\cos x}=1.
Therefore, we have
\begin{align*}
\lim_{x\to 0}\frac{\tan x-x}{x^3}
=&\ \lim_{x\to 0}\frac{\sec^2 x-1}{3x^2}\\
=&\ \lim_{x\to 0}\frac{\tan^2 x}{3x^2}\\
=&\ \frac{1}{3}\lim_{x\to 0}\Big(\frac{\tan x}{x}\Big)^2\\
=&\ \frac{1}{3}\Big(\lim_{x\to 0}\frac{\tan x}{x}\Big)^2\\
\text{ Use \eqref{eq:30-2} }=&\ \frac{1}{3}\cdot 1=\frac{1}{3}.
\end{align*}

### Part c

We can apply L’Hospital’s Rule repeatedly to get the answer. Please check the conditions for applying L’Hospital’s Rule carefully for every step. We have
\begin{align*}
\lim_{x\to 0}\left(\frac{1}{\sin x}-\frac{1}{x}\right)
=&\ \lim_{x\to 0}\frac{x-\sin x}{x\sin x}\\
=&\ \lim_{x\to 0}\frac{1-\cos x}{\sin x+x\cos x}\\
=&\ \lim_{x\to 0}\frac{\sin x}{\cos x+\cos x-x\sin x}\\
=&\ \frac{0}{1+1-0}=0.
\end{align*}

### Part d

We have
$$\ln(\cos x)^{1/x^2}=\frac{1}{x^2}\ln(\cos x).$$ Instead of computing $\lim_{x\to 0} (\cos x)^{1/x^2}$ directly, we compute $\lim_{x\to 0}\dfrac{1}{x^2}\ln(\cos x)$ first. We can apply L’Hospital’s Rule and obtain that
\begin{align*}
\lim_{x\to 0}\frac{1}{x^2}\ln(\cos x)
=&\ \lim_{x\to 0}\frac{\ln(\cos x)}{x^2}\\
=&\ \lim_{x\to 0}\frac{ \frac{-\sin x}{\cos x} }{2x}\\
=&\ -\frac{1}{2}\lim_{x\to 0}\frac{\tan x}{x}\\
\text{ Use \eqref{eq:30-2} }=&\ -\frac{1}{2}.
\end{align*} Therefore, by Theorem 20.5, we have
$$\lim_{x\to 0}\ln(\cos x)^{1/x^2}=e^{-1/2}=\frac{1}{\sqrt e}.$$
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://linearalgebras.com/elementary-analysis-30-02.html", "fetch_time": 1709270123000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-10/segments/1707947474948.91/warc/CC-MAIN-20240301030138-20240301060138-00711.warc.gz", "warc_record_offset": 359763772, "warc_record_length": 20503, "token_count": 915, "char_count": 2091, "score": 4.8125, "int_score": 5, "crawl": "CC-MAIN-2024-10", "snapshot_type": "longest", "language": "en", "language_score": 0.49917253851890564, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00027-of-00064.parquet"}