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limit as x approach… - QuestionCove
OpenStudy (anonymous):

limit as x approaches 0 of sin(x)/x^2-x

4 years ago
OpenStudy (anonymous):

limit as x approaches 0 for: \[\sin (x) \div x^2-x\]

4 years ago
OpenStudy (anonymous):

Use L'Hopital's rule. Are you familiar with that?

4 years ago
OpenStudy (anonymous):

i think i've heard it before, however, i don't know what it is

4 years ago
OpenStudy (anonymous):

If you have a limit where both the numerator and denominator = 0 when you plug in the number your limit is approaching, then you can take the derivative of the numerator and the derivative of the denominator separately and try plugging in the number your limit is approaching again.

4 years ago
OpenStudy (anonymous):

i'm still in pre-calculus and we haven't learnt derivatives yet...

4 years ago
OpenStudy (anonymous):

Then this problem is ugly. Sorry can't help.

4 years ago
OpenStudy (anonymous):

haha it's fine, thanks for trying

4 years ago
OpenStudy (anonymous):

\[\lim_{x \rightarrow 0}\frac{ \sin x }{ x ^{2} }-x \text{, like this?}\]

4 years ago
OpenStudy (anonymous):

nope, the -x is in the denominator, next to \[x^2\]

4 years ago
OpenStudy (anonymous):

so it's \[x^2-x\] in the denominator

4 years ago
OpenStudy (anonymous):

okay... then factor \[\lim_{x \rightarrow 0}\frac{ \sin x }{ x ^{2}-x }=\lim_{x \rightarrow 0}\frac{ \sin x }{ x \left( x-1 \right) }=\lim_{x \rightarrow 0}\frac{ \sin x }{ x }\frac{ 1 }{ \left( x-1 \right) }\] sin x/x goes to 1 and 1/(x-1) goes to -1 so take the product of the 2

4 years ago
OpenStudy (anonymous):

oh! thank you so much! this makes sense now, i never thought to factor it! thanks :)

4 years ago
OpenStudy (anonymous):

you're welcome!!!

4 years ago
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