Compute lim (x→0) sin(5x)/ x without using L'hopital rule
- Thread starter hndalama
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D
Deleted member 4993
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Use the fact that
\(\displaystyle \lim_{x \to 0}\left[\dfrac{sin(x)}{x}\right] \ = \ 1\)
but what do i do with the 5x that's inside the sin?
D
Deleted member 4993
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Use the fact that
\(\displaystyle \lim_{x\to 0}\left [\dfrac{sin(5x)}{5x}\right ] \ = \ 1\)
Steven G
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If lim (x→0) sin(x)/ x =1, then lim (x→0) sin(14x)/ (14x) = 1, then lim (x→2) sin(x-2)/ (x-2) = 1 and lim (x→00) sin(1/x)/ (1/x) =1
Two things need to happen to get 1
1) the angle (what you are taking sin of) and what you are dividing by must be the same.
and 2) what x is approaching should make the angle approach 0.
Also note that lim(x→0) x/sin(x) =1
Bonus: \(\displaystyle lim(x→0) \dfrac{sin(ax)}{bx} = lim(x→0) \dfrac{a}{a}\dfrac{sin(ax)}{bx}= lim(x→0) \dfrac{a}{b}\dfrac{sin(ax)}{ax}\)
\(\displaystyle =\dfrac{a}{b}lim(x→0) \dfrac{sin(ax)}{ax}=(\dfrac{a}{b})1 \ \ =\dfrac{a}{b}\)
Do you see that you can read off the answer?? Do you see how you can use the above factto read off the answer to your problem??
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