Question

$\bf\red{QƲƐŞƬiøƝ}$
$\bf{\displaystyle\lim_{y \to 0}\rm \frac{(x + y).sec(x + y) - xsecx}{ {y}}}$

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Answer 1

Step-by-step explanation:

The value of

y→0

lim

y

(x+y)sec(x+y)−xsecx

is equal to

à sec square

Answer 2

Answer:

$\bf\red{QƲƐŞƬiøƝ}$⠀⠀⠀

$\bf{\displaystyle\lim_{y \to 0}\rm \frac{(x + y).sec(x + y) - xsecx}{ {y}}}$⠀⠀⠀

⠀

$\bf\red{AƝŞᗯƐR}$⠀

$\bf{↦ \: \displaystyle\lim_{y \to 0}\rm \ \frac{x \: sec(x + y) + y \: sec(x + y) - x \: secx}{y} }$

⠀

$\bf{↦ \: \displaystyle\lim_{y \to 0}\rm \ \frac{x \:[ sec(x + y) - sec \: x] + y \: sec(x + y)}{y} }$⠀⠀

⠀

$\bf{↦ \: \displaystyle\lim_{y \to 0}\rm \ \frac{x}{y} \frac{ \:[ \frac{1}{cos(x + y)} - \frac{1}{cos \: x} ] + \frac{y}{y} [ \frac{1}{cos(x + y)} ]}{y} }$⠀⠀

⠀

$\bf{↦ \: \displaystyle\lim_{y \to 0}\rm \ \: \: \frac{x}{y} \frac{ \:[ \frac{cos - cs(x + y)}{cos(x + y)} ] + [ \frac{1}{cos(x + y)} ]}{y} }$⠀⠀

⠀

$\bf{↦ \: \displaystyle\lim_{y \to 0}\rm \ \: \: \frac{x}{y} \frac{ \:[ 2 \: sin \: \frac{x + x + y}{2} . \: sin \ \frac{x + y - x}{2} ] + [ \frac{1}{cos(x + y)} ]}{y} }$⠀⠀

⠀

$\bf{↦ \: \displaystyle\lim_{y \to 0}\rm \ \: \: \frac{x}{y} \frac{ \:[ \frac{sin \: 2x + y}{2} . \: sin \frac{y}{2} ] + [ \frac{1}{cos(x + y)} ]}{y} }$⠀⠀

⠀

$\bf{↦ \: 2x \: \: \frac{\displaystyle\lim_{y \to 0}\rm \ \: \: \frac{sin(x + y)}{2} \: \displaystyle\lim_{y \to 0}\rm \ \: sin \frac{y}{2} }⠀⠀</p><p></p><p>⠀{\displaystyle\lim_{y \to 0}\rm \ \: cos(x + y) \: \displaystyle\lim_{y \to 0}\rm \ \: cos \: x} \: \: + \: \: \displaystyle\lim_{y \to 0}\rm \ \: \frac{1}{cos(x + y)} }$⠀⠀

⠀

$\bf{↦ \: 2x \: \: \frac{\displaystyle\lim_{y \to 0}\rm \ \: \: \frac{sin(x + y)}{2} \: \displaystyle\lim_{y \to 0}\rm \ \: sin \ \frac{ \frac{y}{2} }{ \frac{y}{2}.2 } }{\displaystyle\lim_{y \to 0}\rm \ \: cos(x + y) \: \displaystyle\lim_{y \to 0}\rm \ \: cos \: x} \: \: + \: \: \displaystyle\lim_{y \to 0}\rm \ \: \frac{1}{cos(x + y)} }$⠀⠀

⠀

$\bf\green{N.T.L}$

⠀⠀⠀⠀⠀⠀⠀Now Taking Limit ⠀⠀⠀⠀⠀⠀⠀

$↦ \: \bf{ \frac{2x \times sinx}{cosx.cosx} \times \frac{1}{2} + \frac{1}{cosx} }$

$↦ \: \bf{x \: tanx. secx + secx}$

$↦ \: \bf{x \: secx. tanx + secx}$

⠀

⠀⠀⠀⠀$➾ \: \bf\pink{ sec x ( x \: tanx +1)}$