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How would I solve the following limits? As $$\lim_{x\to 0}x^2(1+ \cot^2{3x})$$ I know $\cot^2(3x)$ is $\displaystyle\frac{\cos^2{3x}}{\sin^2{3x}}$ but I am not sure how to solve it. The second question is how would I solve $$\lim_{x\to\pi/4}\frac{1-\cos x}{x}$$ |
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If I am reading your question right, we want $$\lim_{x\to 0}x^2(1+\cot^2 3x).$$ How to approach? A reasonable thing to do is to express the cotangent in terms of the more familiar sines and cosines: $\cot t=\dfrac{\cos t}{\sin t}$. A little bit of algebra shows that we want $$\lim_{x\to 0} \frac{x^2}{\sin^2 3x}(1+\cos^2 3x).$$ The $1+\cos^2 3x$ part is "safe," it approaches $2$. To deal with the rest, express it as $\dfrac{1}{9}\left(\dfrac{3x}{\sin 3x}\right)^2$. Now you can take over. In the second question, we want $$\lim_{x\to\pi/4}\frac{1-\cos x}{x}.$$ This one requires almost no work. As $x$ approaches $\pi/4$, top and bottom behave perfectly nicely. The top approaches $1-\cos(\pi/4)$, since cosine is continuous. And of course the bottom approaches $\pi/4$. For a more explicit answer, recall that $\cos(\pi/4)=1/\sqrt{2}$. So the limit is equal to $$\frac{1-\frac{1}{\sqrt{2}}}{\pi/4}.$$ |
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Recall that $1+\cot^2(y) = \csc^2(y)$. Hence, $$\lim_{x \to 0} x^2 (1+\cot^2(3x)) = \lim_{x \to 0} x^2 \csc^2(3x) = \lim_{x \to 0} \left(\dfrac{x}{\sin(3x)} \right)^2 = \dfrac19$$ For the second problem, as $x \to \pi/4$, the numerator tends to $1-\dfrac1{\sqrt{2}}$, while the denominator tends to $\pi/4$. Hence, the limit is $$\dfrac{1-\dfrac1{\sqrt{2}}}{\pi/4} = \dfrac{4-2\sqrt2}{\pi}$$ |
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