Question () Watch Video Show Evaluate the limit and express your answer in simplest form. \[ \lim _{h \rightarrow 0} \frac{\cos ^{-1}\left(\frac{3}{5}+h\right)-\cos ^{-1}\left(\frac{3}{5}\right)}{h} \] Answer Attempt 1 out of 3 Submit Answer

Step 1 ：Simplify the expression inside the limit

Step 2 ：\(\frac{\cos^{-1}\left(\frac{3}{5}+h\right)-\cos^{-1}\left(\frac{3}{5}\right)}{h}\)

Step 3 ：Apply the identity \(\cos^{-1}(a) - \cos^{-1}(b) = \cos^{-1}(a\sqrt{1-b^2} + b\sqrt{1-a^2})\)

Step 4 ：\(\frac{\cos^{-1}\left(\frac{3}{5}\sqrt{1-\left(\frac{3}{5}+h\right)^2}+\left(\frac{3}{5}+h\right)\sqrt{1-\left(\frac{3}{5}\right)^2}\right)}{h}\)

Step 5 ：Simplify the expression inside the inverse cosine function

Step 6 ：\(\frac{\cos^{-1}\left(\frac{3}{5}\sqrt{\frac{16}{25}-\frac{6h}{5}-h^2}+\frac{3}{5}\sqrt{\frac{16}{25}}+\frac{3}{5}h\right)}{h}\)

Step 7 ：Evaluate the limit as \(h\) approaches 0

Step 8 ：\(\lim _{h \rightarrow 0} \frac{\cos^{-1}\left(\frac{3}{5}\sqrt{\frac{16}{25}-\frac{6h}{5}-h^2}+\frac{3}{5}\sqrt{\frac{16}{25}}+\frac{3}{5}h\right)}{h}\)

Step 9 ：Apply the derivative of the inverse cosine function

Step 10 ：\(\lim _{h \rightarrow 0} -\frac{1}{\sqrt{1-\left(\frac{3}{5}\sqrt{\frac{16}{25}-\frac{6h}{5}-h^2}+\frac{3}{5}\sqrt{\frac{16}{25}}+\frac{3}{5}h\right)^2}}\)

Step 11 ：Evaluate the limit as \(h\) approaches 0

Step 12 ：\(\lim _{h \rightarrow 0} -\frac{1}{\sqrt{1-\left(\frac{3}{5}\sqrt{\frac{16}{25}}\right)^2}}\)

Step 13 ：Simplify the expression

Step 14 ：\(\lim _{h \rightarrow 0} -\frac{1}{\sqrt{1-\frac{9}{25}}}\)

Step 15 ：Evaluate the limit

Step 16 ：\(\lim _{h \rightarrow 0} -\frac{1}{\frac{4}{5}}\)

Step 17 ：Simplify the expression

Step 18 ：\(\lim _{h \rightarrow 0} -\frac{5}{4}\)

Step 19 ：Final Answer: \(\boxed{-\frac{5}{4}}\)