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प्रश्न
Using L' Hospital's rule, evaluate:
`lim_("x"→0) (1/"x"^2 - cot"x"/"x")`
उत्तर
`lim_("x"→0) (1/"x"^2-cot"x"/"x")`
`= lim_("x"→0) (1/"x"^2-1/("x"tan "x"))`
` = lim_("x"→0) (tan "x"- "x")/("x"^2 tan "x")`
` = lim_("x"→0) (tan "x"-"x")/"x"^3 . ("x"/tan"x")`
` = (lim_("x"→0) (tan"x"-"x")/"x"^3) (lim_("x"→0) "x"/(tan"x"))`
` = (lim_("x"→0) (tan"x"-"x")/"x"^3) . 1`
` = lim_("x"→0) (sec^2 "x"-1)/(3"x"^2) (0/0)`
` = (lim_("x"→0) (sec^2"x")/3) (lim_("x"→0)tan"x"/"x")`
` = lim_("x"→0) sec^2"x"/3 . 1 . = 1/3`
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