lim(x->0) tan^2(αx)*sin(βx)/x^3
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pka
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the teacher gave us this as a homework , and i tried really hard to solve it but I couldn't
so I really need help
lim(x->0) tan^2(αx)*sin(βx)/x^3
Rewrite as \(\displaystyle (\alpha)^2\dfrac{\sin^2(\alpha x)}{(\alpha x)^2}\)\(\displaystyle ~ \dfrac{1}{\cos^2(\alpha x)}~\beta \dfrac{sin(\beta x)}{\beta x}\)
What are you allowed to use? For example, if you were allowed to use
\(\displaystyle sin(\beta x) = \beta x - \frac{(\beta x)^3}{6} + ... = \beta x [ 1 - \frac{(\beta x)^2}{6} + ...] \)
\(\displaystyle tan(\alpha x) = \alpha x + \frac{(\alpha x)^3}{3} + ... = \alpha x [ 1 + \frac{(\alpha x)^2}{3} + ...]\)
then we would have
=\(\displaystyle lim_{x \to 0}\frac{\beta x [ 1 - \frac{(\beta x)^2}{6} + ...] (\alpha x)^2 [ 1 + \frac{(\alpha x)^2}{3} + ...]^2}{x^3}\)
=\(\displaystyle lim_{x \to 0}{\beta (\alpha)^2 [ 1 - \frac{(\beta x)^2}{6} + ...] [ 1 + \frac{(\alpha x)^2}{3} + ...]^2}\)
=\(\displaystyle \beta {\alpha}^2\)
If you are only allowed to use something like L'hospital's rule, you will need to plow your way through.
Note: If you were allowed to use
\(\displaystyle lim_{x \to 0}\frac{sin(x)}{x} = 1\)
the same sort of limit could be done. Hint: let u = \(\displaystyle \beta x\) and v = \(\displaystyle \alpha x\)
\(\displaystyle sin(\beta x) = \beta x - \frac{(\beta x)^3}{6} + ... = \beta x [ 1 - \frac{(\beta x)^2}{6} + ...] \)
\(\displaystyle tan(\alpha x) = \alpha x + \frac{(\alpha x)^3}{3} + ... = \alpha x [ 1 + \frac{(\alpha x)^2}{3} + ...]\)
then we would have
=\(\displaystyle lim_{x \to 0}\frac{\beta x [ 1 - \frac{(\beta x)^2}{6} + ...] (\alpha x)^2 [ 1 + \frac{(\alpha x)^2}{3} + ...]^2}{x^3}\)
=\(\displaystyle lim_{x \to 0}{\beta (\alpha)^2 [ 1 - \frac{(\beta x)^2}{6} + ...] [ 1 + \frac{(\alpha x)^2}{3} + ...]^2}\)
=\(\displaystyle \beta {\alpha}^2\)
If you are only allowed to use something like L'hospital's rule, you will need to plow your way through.
Note: If you were allowed to use
\(\displaystyle lim_{x \to 0}\frac{sin(x)}{x} = 1\)
the same sort of limit could be done. Hint: let u = \(\displaystyle \beta x\) and v = \(\displaystyle \alpha x\)
I am only allowed to use
[FONT=MathJax_Math-italic]l[/FONT][FONT=MathJax_Math-italic]i[/FONT][FONT=MathJax_Math-italic]m (x->0) [/FONT][FONT=MathJax_Math-italic]s[/FONT][FONT=MathJax_Math-italic]i[/FONT][FONT=MathJax_Math-italic]n[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main])/[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]1 , and [/FONT][FONT=MathJax_Math-italic]l[/FONT][FONT=MathJax_Math-italic]i[/FONT][FONT=MathJax_Math-italic]m (x->0) tan[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main])/[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]1[/FONT]
so I tried to solve it like this
there is only the cos left , I think it'll be 1/1=1 but im not sure if it right ?
[FONT=MathJax_Math-italic]l[/FONT][FONT=MathJax_Math-italic]i[/FONT][FONT=MathJax_Math-italic]m (x->0) [/FONT][FONT=MathJax_Math-italic]s[/FONT][FONT=MathJax_Math-italic]i[/FONT][FONT=MathJax_Math-italic]n[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main])/[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]1 , and [/FONT][FONT=MathJax_Math-italic]l[/FONT][FONT=MathJax_Math-italic]i[/FONT][FONT=MathJax_Math-italic]m (x->0) tan[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main])/[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]1[/FONT]
so I tried to solve it like this
there is only the cos left , I think it'll be 1/1=1 but im not sure if it right ?
Last edited:
Yes that is correct.