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Greg C.
Ph.D. in Mathematics; 10 years tutor experience
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Pre-Calculus
TutorMe
Question:

Simplify the trigonometric expression: $$\tan^2 x \cot x \csc x$$

Greg C.

A common strategy to simplify complicated trig expressions like this is to write everything in terms of sin or cos. In this problem, we have $$\tan x = \frac{\sin x}{\cos x}$$ and so $$\tan^2x = \frac{\sin^2 x}{\cos^2 x}$$ $$\cot x = \frac{\cos x}{\sin x}$$ and $$\csc x = \frac{1}{\sin x}$$ Thus, we have $$\tan^2 x \cot x \csc x = \frac{\sin^2 x}{\cos^2 x} \cdot \frac{\cos x}{\sin x} \cdot \frac{1}{\sin x}$$ which, after simplifying the fractions, equals $$\frac{1}{\cos x} = \sec x$$ Thus, $$\tan^2 x \cot x \csc x = \sec x$$

Calculus
TutorMe
Question:

Find the limit: $$\lim_{x \rightarrow 0^+} (1+4x)^{1/x}$$

Greg C.

$$1 + 4x \rightarrow 1$$ as $$x \rightarrow 0^+$$, and $$1/x \rightarrow \infty$$ as $$x \rightarrow 0^+$$, so this looks like a $$1 ^ \infty$$ type problem. Unfortunately, $$1 ^\infty$$ is an "indeterminate form", meaning we have to do some extra work to figure out what the answer is (don't assume the answer is 1!) Many inderterminate form limit problems are solved with L'Hospital's rule; but at this moment we cannot use L'Hospital's rule because L'Hospitals rule only applies for limits of the form 0/0 or $$\infty/\infty$$. So to solve the problem, we will do the following trick: let $$y = \lim_{x \rightarrow 0^+} (1+4x)^{1/x}$$. Take the natural log of both sides. $$\implies \ln y = \lim_{x \rightarrow 0^+} \ln (1+4x)^{1/x^2}$$ (note that ln lim = lim ln. You can switch them ) Recall that with logs, we can "bring the exponent down", (the power rule for logs) that is, $$\ln (1+4x)^{1/x} = \frac{1}{x} \cdot \ln(1+4x) = \frac{\ln (1+4x)}{x}$$ Thus, $$\lim_{x \rightarrow 0^+} \ln (1+4x)^{1/x} = \lim_{x \rightarrow 0^+} \frac{\ln (1+4x)}{x} = \frac{0}{0}$$ an indeterminate form (recall $$\ln 1 = 0$$). But here's the point: For the indeterminate form $$\frac{0}{0}$$, we can use L'Hospital's rule! That's the whole point of all this natural log stuff. We were trying to rewrite the problem in such a way that we can use L'Hospital's rule. So, with L'Hospital's rule, we have $$\lim_{x \rightarrow 0^+} \frac{\ln (1+4x)}{x} = \lim_{x \rightarrow 0^+} \frac{ \frac{4}{1+4x}}{1} = \lim_{x \rightarrow 0^+} \frac{4}{1+4x} = 4$$ Thus, we (finally) have $$\ln y = \lim_{x \rightarrow 0^+} \ln (1+4x)^{1/x} = 4$$ So $$\ln y = 4$$ and so $$y = e^4$$. But remember, $$y = \lim_{x \rightarrow 0^+} (1+4x)^{1/x}$$, so our final answer is $$e^4$$. It is common with limit problems of the form $$1^\infty$$ to contain $$e$$ somewhere in the answer. (of course this is not always the case)

Algebra
TutorMe
Question:

Let $$f(x) = \frac{x^2 -4}{x -2}$$ and $$g(x) = x + 2.$$ Does f(x) = g(x)?

Greg C.

No, because f(2) is undefined while g(2) =4. Indeed, if we plug 2 into f(x), we will have f(2) = 0/0, which is undefined (you can NEVER have 0 on the bottom in the denominator). But we can legally plug 2 into g(x) and have g(2) = 4. Thus, 2 is NOT in the domain of f, but 2 IS in the domain of g. That is, the domain of f is $$(-\infty, 2) \cup (2, \infty)$$ while the domin of g is all real numbers, that is, $$(-\infty, \infty)$$ A common error is to write $$f(x) = \frac{x^2 -4}{x -2} = \frac{(x-2)(x+2)}{x-2} = x +2$$ and so many students think f(x) = g(x). But they disagree on x=2, and therefore have different domains, and therefore are NOT equal functions.

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