Find the limit of (1/sinx - 1/tanx) as x approaches 0 WITHOUT L'Hopitals Rule!

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In this video we find the limit of of (1/sinx - 1/tanx) as x approaches 0 without using L'Hopital's rule. Each step is explained carefully. Limit laws, trig identities etc are all taken for granted in this problem.

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As an alternative, you can use tangent half angle identities for the sine and the tangent, which are

sin(x) = 2·tan(½x)/(1 + tan²(½x))
tan(x) = 2·tan(½x)/(1 − tan²(½x))

These and related identities are used in integral calculus to express the trigonometric functions sin(x) = 2t/(1 + t²), cos(x) = (1 − t²)/(1 + t²), tan(x) = 2t/(1 − t²) as rational functions of tan(½x) = t. From these identities we have

1/sin(x) − 1/tan(x) = (1 + tan²(½x))/(2·tan(½x)) − (1 − tan²(½x))/(2·tan(½x))

which simplifies to

1/sin(x) − 1/tan(x) = 2·tan²(½x)/(2·tan(½x)) = tan(½x)

Consequently

lim ₓ → ₀ (1/sin(x) − 1/tan(x)) = lim ₓ → ₀ tan(½x) = 0

NadiehFan

Just use small angle approximations. When x is small, sin(x)≈x
tan(x)≈x
So as x->0,
Sin(x)=x
Tan(x)=x
So
Lim(x->0) (1/sin(x) - 1/tan(x))
Transforms to
Lim(x->0) (1/x - 1/x) = 0

harryhaste

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