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lim⁡x,y,z→3,3,0 sin⁡2x+cos2y+sec2z\lim_{x,y,z\to3,3,0}\:\sin^2x+cos^2y+sec^2z

Step-by-step Solution

1

Evaluate the limit lim⁡z→3.3,0(sin⁡(x)2+cos⁡(y)2+sec⁡(z)2)\lim_{z\to{3.3,0}}\left(\sin\left(x\right)^2+\cos\left(y\right)^2+\sec\left(z\right)^2\right) by replacing all occurrences of zz by 3.3,03.3,0

x,y,sin⁡(x)2+cos⁡(y)2+sec⁡(3.3,0)2x,y,\sin\left(x\right)^2+\cos\left(y\right)^2+\sec\left(3.3,0\right)^2

Final answer to the exercise

x,y,sin⁡(x)2+cos⁡(y)2+sec⁡(3.3,0)2x,y,\sin\left(x\right)^2+\cos\left(y\right)^2+\sec\left(3.3,0\right)^2

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Main Topic: Limits

In mathematics, a limit is the value that a function or sequence "approaches" as the input or index approaches some value.

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limx,y,z→3,3,0 sin2x+cos2y+sec2z
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