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Tutor profile: Ben Z.

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Ben Z.
Attentitive & Passionate Ivy League STEM Tutor
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Questions

Subject: Pre-Calculus

TutorMe
Question:

Describe the graph of $$f(x)=\frac{(x-2)}{(x-2)}$$.

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Ben Z.
Answer:

There's a few things we might be tempted to do, that aren't right. We can't go straight to graphing asymptotes and roots. If we set the numerator equal to 0, we get $$x=2$$ is a root of the function. If we set the denominator equal to 0, we get $$x=2$$ is an asymptote of the function. This doesn't make sense. We can't hastily cancel out the $$x-2$$ from top and bottom. There's a small problem with this: what happens if $$x=2$$?? Instead, we notice we can cancel out the $$x-2$$ ONLY IF $$x\neq2$$. Therefore, $$f(x)=1$$ whenever $$x\neq 2$$. But, when $$x=2$$, $$f(x)$$ is undefined, so we have a hole in the graph at that point. Our graph looks like a horizontal line with a small hole at the point $$(2,1)$$.

Subject: Calculus

TutorMe
Question:

Given a positive integer $$n$$, evaluate $$\lim\limits_{x\to\infty}\frac{x^n}{e^x}$$.

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Ben Z.
Answer:

How can we possibly evaluate this problem when we don't know what $$n$$ is? Try small cases. Plug in $$n=1$$ to see we need to find $$\lim\limits_{x\to\infty}\frac{x}{e^x}$$. We can use L'Hopital's rule to find this limit, by taking the derivatives of top and bottom and getting $$\lim\limits_{x\to\infty}\frac{1}{e^x}=0$$. Maybe this technique will help with the problem at hand: $$\lim\limits_{x\to\infty}\frac{x^n}{e^x}=\lim\limits_{x\to\infty}\frac{nx^{n-1}}{e^x}=\lim\limits_{x\to\infty}\frac{n(n-1)x^{n-2}}{e^x}...$$ We can start to see if we apply L'Hopital's rule $$n+1$$ times, the numerator will disappear! This is because if we take enough derivatives of $$x^n$$, we'll get 0 after a while. Because $$\frac{d}{dx}e^x=e^x$$, no matter how many times we apply L'Hopital's rule, we'll still have $$e^x$$ in the bottom. Therefore, no matter how big $$n$$ is, the limit is 0!

Subject: Physics

TutorMe
Question:

A ball is dropped from the top of a huge building with height $$H$$ at the same time another ball is thrown upwards from the ground with a speed $$v$$. After what amount time do the balls meet each other?

Inactive
Ben Z.
Answer:

We could set up two different equations for the positions of the balls in terms of $$g$$, $$t$$, and $$v$$, and solve it that way. Or, we can consider the problem from the frame of reference of the ball that was originally on the ground. Both the balls are accelerating downwards at a rate of $$g$$, so there is no relative acceleration! The bottom ball is approaching the top ball at a speed of $$v$$, and they are a distance $$H$$ apart, so using the equation $$d=vt$$ we get $$vt=H$$ , and therefore, $$t=\frac{H}{v}$$ is the solution.

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