Question

How do you find \[\lim \dfrac{{\cos x}}{x}\] as \[x \to {0^ + }\]?

Answer 1

Hint: In the given question, we have been given to find the value of the limit of a function of cosine divided by its argument as its argument approached zero plus. We are going to use the value of the Maclaurin expansion of cosine. Then we are going to put the value into the limit, simplify the value of the expansion, and then plug in the value of the argument and find our answer.

Formula Used:
We are going to use the formula of Maclaurin expansion of \[\cos x\], which is
\[\cos x = 1 - \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^4}}}{{4!}} - \dfrac{{{x^6}}}{{6!}} + ...\]

Complete step-by-step answer:
The given expression is:
\[{\lim _{x \to {0^ + }}}\dfrac{{\cos x}}{x}\]
First, let us simplify the expression \[\dfrac{{\cos x}}{x}\] by using Maclaurin expansion of \[\cos x\], which is
\[\cos x = 1 - \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^4}}}{{4!}} - \dfrac{{{x^6}}}{{6!}} + ...\]
Now, putting the value into \[\dfrac{{\cos x}}{x}\], and we get,
\[\dfrac{{\cos x}}{x} = \dfrac{{1 - \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^4}}}{{4!}} - \dfrac{{{x^6}}}{{6!}} + ...}}{x} = \dfrac{1}{x} - \dfrac{{{x^1}}}{{2!}} + \dfrac{{{x^3}}}{{4!}} - \dfrac{{{x^5}}}{{6!}} + ...\]
As, \[x \to {0^ + }\], all the terms after and including \[\dfrac{x}{{2!}}\] become \[0\].
Hence, we are left with,
\[{\lim _{x \to {0^ + }}}\dfrac{1}{x}\]
Now, putting in the value to evaluate the limit, we get,
\[{\lim _{x \to {0^ + }}}\dfrac{1}{x} = \dfrac{1}{0} = \infty \]

Additional Information:
We could have just solved the answer by putting in the value of \[x\]:
\[{\lim _{x \to {0^ + }}}\dfrac{{\cos x}}{x} = \dfrac{1}{0} = \infty \]
but we applied the available formula to show that there is not any other option available for the value.

Note: In this question, we had to find the value of the limit of a function of cosine divided by its argument as its argument approached zero plus. We could have just plugged in the value of the argument and we could have just gotten the answer. But we applied a more sophisticated, more describing approach so as to convert the limit into the simplest form and then we applied the argument and found the answer.