Hi, there!
On Day One, when I joined a trading company, fresh out of college,
I might have looked a bit tense because I thought I needed to show
off my specific command of English in either speaking or writing.
Also, I might have had a haunted look in my eyes, for I was unsure
whether I could become a welcome addition to the export section
I joined. My seniors told me, “Just holler if you have any questions.”
I wrung out my voice in responding to their offers by saying, “Thank
you.”
After lunch on that day, what was I doing? Typing? Answering a pho-
ne call? … No, not that kind of job the novice employees usually do.
I kicked off work by calculating several quotations on Day One. I ne-
ver ever imagined I would live out of a suitcase a few years later.
Thanks to the inception of the calculation job, I have become calcu-
lative. Thus, getting the drift above, let me try the two estimations.
Evaluate each of the following limits:
(1) \(\displaystyle\lim_{x\rightarrow 0}(\frac{sin\,x}{x})\)
(2) \(\displaystyle\lim_{t\rightarrow 1}(\frac{4t^{5} – 2t^{4} – 2}{5 – t – 4t^{4}})\)
Here are my solutions to the questions above.
(1) This is a \(\frac{0}{0}\) indeterminate form, so let me apply L’Hopital’s Rule.
\(\displaystyle\lim_{x\rightarrow 0}(\frac{sin\,x}{x}) = \displaystyle\lim_{x\rightarrow 0}(\frac{cos\,x}{1}) = \frac{1}{1} = 1\)
(2) In this case, this also has a \(\frac{0}{0}\) indeterminate form, and I could
factor the numerator and denominator, simplify, and take the limit.
\(\displaystyle\lim_{t\rightarrow 1}(\frac{4t^{5} – 2t^{4} – 2}{5 – t – 4t^{4}})\)
\(= \displaystyle\lim_{t\rightarrow 1}(\frac{20t^{4} – 8t^{3}}{-1 -16t^{3}})\)
\(= \frac{20 – 8}{-1 – 16}\)
\(= -\frac{12}{17}\)
※L’Hopital’s Rule = a theorem which provides a technique to evaluate
limits of indeterminate forms
Stay tuned, and expect to see my next post.
Keep well.
Frank Yoshida
￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣
【グローバルビジネスで役立つ数学】でもっと学習する b＾＾)
【参考図書】『もう一度高校数学』（著者：高橋一雄氏）株式会社日本実業出版社
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