Math Challenges

Submissions for Problem #3

Problem #3

Evaluate the following limits

\[ \displaylines{\lim_{t\rightarrow0}\frac{\sin\left(\pi t\right)}{t}\ \ \\ \\ \lim_{x\rightarrow3}\frac{x^2-9}{x-3}\ \ \\ \\ \lim_{k\rightarrow-2}\frac{k+2}{k^2-4}} \]

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fish-face

Solution: \( \displaylines{\pi\\ 6\\ -\frac14} \)

Explanation:
1. Use L'Hopital and evaluate the limit of pi cos(pi*t) / 1, which is continuous and equal to pi at 0. 2. This function is defined and equal to x + 3 everywhere except 3, and since this is continuous, we can evaluate it at 3 to get the limit of the original function. 3. Similar to 2, we can evaluate the function 1/(k-2) which is defined and continuous on an interval around -2, with value -1/4.

zapwai

Solution: \( \displaylines{\text{ Given}f\left(t\right)=\frac{\sin\left(\pi t\right)}{t}\text{ we have}\\ t\to f\left(t\right)\\ .1\to3.09016994374947\\ .01\to3.14107590781283\\ .001\to3.14158748587956\\ .0001\to3.14159260191266\\ .00001\to3.14159265307302\\ \\ \\ \frac{x^2-9}{x-3}=\frac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)}=\left(x+3\right)\Rightarrow\lim_{x\to3}\frac{x^2-9}{x-3}=\lim_{x\to3}x+3=6\\ \\ \frac{k+2}{k^2-4}=\frac{\left(k+2\right)}{\left(k+2\right)\left(k-2\right)}=\frac{1}{k-2}\Rightarrow\lim_{k\to-2}\frac{k+2}{k^2-4}=\lim_{k\to-2}\frac{1}{k-2}=\frac{1}{-4}} \)

Explanation:
For the first question, it is common in a beginning calculus course (when you don't know L'Hopital's rule, for example) to simply make a table with values of t and f(t). Here we take values of t approaching 0 from the right and see that f(t) tends to the value pi. (One should also check values from the left, e.g. f(-0.00001) = 3.14159265307302) The second and third problems can be factored and reveal a removable discontinuity, allowing us to simply evaluate at the given value (because the resulting functions are continuous).