anonymous
  • anonymous
if lim(x-->0)f(x)=-1find the lim(x-->0) (f(x)+1)sin(pi/x^2)
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
jtvatsim
  • jtvatsim
Could we use the squeeze theorem on this one?
jtvatsim
  • jtvatsim
Consider that \[(f(x) + 1) \cdot -1 \leq (f(x) + 1) \cdot \sin(\pi/x^2) \leq (f(x) + 1) \cdot 1\] hence, \[\lim_{x\rightarrow0}(f(x) + 1) \cdot -1 \leq \lim_{x\rightarrow0}(f(x) + 1) \cdot \sin(\pi/x^2) \leq \lim_{x\rightarrow0}(f(x) + 1) \cdot 1\] that is, \[ 0\leq \lim_{x\rightarrow0}(f(x)+1) \cdot \sin(\pi/x^2) \leq 0\] and our desired limit is simply 0.
jtvatsim
  • jtvatsim
The sin(pi/x^2) feels uncomfortable since we "know" that the inside is heading toward something undefined (or infinite), but the sin function itself is always stuck between -1 and 1 no matter what the inside is.

Looking for something else?

Not the answer you are looking for? Search for more explanations.