Limit of x*tan(1/x) as x approaches Infinity (no L'Hospital's Rule) | Calculus 1 Exercises

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We evaluate the limit of xtan(1/x) as x goes to infinity, which gives us a horizontal asymptote of x*tan(1/x). We'll evaluate this limit without using L'Hospital's rule. Instead, we use the limit product rule, the definition of tangent, the trigonometric limit sin(x)/x as x approaches 0, and then we'll pretty much be done showing that the limit as x approaches infinity of x*tan(1/x) is 1! #Calculus1

Limit of sin(x)/x as x approaches 0: (coming soon)

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this was beautiful... it's beautiful

mokoepa

Maybe you can try proving the Bolzano Weierstrass theorem with nested intervals

krasimirronkov

Great a video after long time why didn't you upload?

aashsyed

Can I check why is there a need for L'Hopital rule in this case? It doesn't look like an indeterminate form. Won't the limit of xsin(1/x) just evaluate to infinity*sin(0)=infinity*0=0, and the limit of cos(1/x) just evaluate to cos(0)=1, so the limit of xtan(1/x) is 0/1 = 0?

シンジ碇-fo

Why is this video 9 minutes long for what takes at most 30 seconds

TheBluePhoenix

A question please, lim x tends to infinity (1/tan(x)) is it undefined?

AhmedMostafa-vgod

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