lim x → 2 , y → 1 sin − 1 ⁡ ( x y − 2 ) tan − 1 ⁡ ( 3 x y − 6 )

Question

lim          x      →      2      ,      y      →      1                          sin                  −          1                    ⁡      (      x      y      −      2      )                      tan                  −          1                    ⁡      (      3      x      y      −      6      )

Eden Bradshaw · Accepted Answer

Since x and y only appear as a product, and

x y \to 2

as

x \to 2

and

y \to 1

, it's equivalent to compute the limit

lim_{t \to 2} \frac{\sin^{- 1} (t - 2)}{\tan^{- 1} (3 t - 6)}

The numerator and denominator tend to zero, and they are differentiable, so we can apply L'Hôpital's rule. The derivatives are respectively

\frac{1}{\sqrt{1 - (t - 2)^{2}}}

and

\frac{3}{1 + (3 t - 6)^{2}}

, and they both have a nonzero limit as

t \to 2

, hence

lim_{t \to 2} \frac{\sin^{- 1} (t - 2)}{\tan^{- 1} (3 t - 6)} = lim_{t \to 2} \frac{1 + (3 t - 6)^{2}}{3 \sqrt{1 - (t - 2)^{2}}} = \frac{1}{3}

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