find the value of limit x tends to y for (secx-secy)/(x-y)
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find the value of limit x tends to y for (secx-secy)/(x-y)
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Step-by-step explanation:
We have,
Step-by-step explanation:
We have,
\begin{gathered}\lim _{x \rarr y} \frac{ \sec(x) - \sec(y) }{x - y} \\\end{gathered}x→ylimx−ysec(x)−sec(y)
let \: \: x - y = hletx−y=h
\begin{gathered}\implies \lim _{h \rarr 0} \frac{ \sec(y + h) - \sec(y) }{h} \\\end{gathered}⟹h→0limhsec(y+h)−sec(y)
\begin{gathered}\implies \lim _{h \rarr 0} \frac{ \cos(y) - \cos(y + h) }{h \cos(y) \cos(y + h) } \\\end{gathered}⟹h→0limhcos(y)cos(y+h)cos(y)−cos(y+h)
\begin{gathered}\implies \lim _{h \rarr 0} \frac{ \cos(y) - \cos(y) \cos(h) + \sin(y) \sin(h) }{h \cos(y) \cos(y + h) } \\\end{gathered}⟹h→0limhcos(y)cos(y+h)cos(y)−cos(y)cos(h)+sin(y)sin(h)
\begin{gathered}\implies \lim _{h \rarr 0} \frac{ \cos(y) (1 - \cos(h) )}{h \cos(y) \cos(y + h) } + \lim _{h \rarr 0}\frac{ \sin(h) \sin(y) }{h \cos(y) \cos(y + h) } \\\end{gathered}⟹h→0limhcos(y)cos(y+h)cos(y)(1−cos(h))+h→0limhcos(y)cos(y+h)sin(h)sin(y)
\begin{gathered}\implies \lim _{h \rarr 0} \frac{1 - \cos(h) }{h \cos(y + h) } +1 \times \frac{ \sin(y) }{ \cos(y) \cos(y + 0) } \\\end{gathered}⟹h→0limhcos(y+h)1−cos(h)+1×cos(y)cos(y+0)sin(y)
\implies \lim _{h \rarr 0} \frac{1 - \cos(h) }{h} \times \frac{1}{ \cos(y) } + \tan(y) \sec(y)⟹limh→0h1−cos(h)×cos(y)1+tan(y)sec(y)
= 0 + \tan(y) \sec(y)=0+tan(y)sec(y)
= \tan(y) \sec(y)=tan(y)sec(y)