Question

Find the value of y from the following

$lim \: x \to \: - y \: \frac{ {x}^{9} + {y}^{9} }{x + y} = 9$

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Answer 1

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: \: \: \displaystyle\lim_{x \: \to \: - \: y}\rm \frac{ {x}^{9} + {y}^{9} }{x + y} = 9$

can be rewritten as

$\rm \: \: \displaystyle\lim_{x \: \to \: - \: y}\rm \frac{ {x}^{9} - {( - y)}^{9} }{x - ( - y)} = 9$

We know,

$\boxed{\tt{ \: \: \displaystyle\lim_{x \to a}\rm \frac{ {x}^{n} - {a}^{n} }{x - a} \: = \: {na}^{n - 1} \: \: }}$

So, using this result, we get

$\rm \: \: \: 9 {( - y)}^{9 - 1} = 9$

$\rm \: \: \: {( - y)}^{8} = 1$

$\rm\implies \: {y}^{8} = 1$

$\rm\implies \:y \: = \: \pm \: 1$

ALTERNATIVE METHOD

Given expression is

$\rm \: \: \: \displaystyle\lim_{x \: \to \: - \: y}\rm \frac{ {x}^{9} + {y}^{9} }{x + y} = 9$

If we substitute directly x = - y, we get

$\rm \: \: \dfrac{ { - y}^{9} + {y}^{9} }{ - y + y} = 9$

$\rm \: \: \: \dfrac{0}{0} = 9$

which is indeterminant form.

Now,

$\rm \: \: \: \displaystyle\lim_{x \: \to \: - \: y}\rm \frac{ {x}^{9} + {y}^{9} }{x + y} = 9$

So, Using L - Hospital Rule, we have

$\rm \: \: \: \displaystyle\lim_{x \: \to \: - \: y}\rm \: \frac{\dfrac{d}{dx} {x}^{9} + \dfrac{d}{dx}{y}^{9} }{\dfrac{d}{dx}x + \dfrac{d}{dx}y} = 9$

We know,

$\boxed{\tt{ \dfrac{d}{dx} {x}^{n} \: = \: {nx}^{n - 1} \: }}$

and

$\boxed{\tt{ \dfrac{d}{dx}k \: = \: 0 \: }}$

So, using these results, we get

$\rm \: \: \displaystyle\lim_{x \: \to \: - y}\rm \frac{ {9x}^{9 - 1} + 0}{1 + 0} = 9$

$\rm \: \: \displaystyle\lim_{x \: \to \: - y}\rm \frac{ {9x}^{8}}{1} = 9$

$\rm \: \: \: {9( - y)}^{8} = 9$

$\rm\implies \: {y}^{8} = 1$

$\rm\implies \:y \: = \: \pm \: 1$

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ADDITIONAL INFORMATION

$\rm \: \: \: \displaystyle\lim_{x \to 0}\rm \: \frac{sinx}{x} \: = \: 1$

$\rm \: \: \: \displaystyle\lim_{x \to 0}\rm \: \frac{tanx}{x} \: = \: 1$

$\rm \: \: \: \displaystyle\lim_{x \to 0}\rm \: \frac{log(1 + x)}{x} \: = \: 1$

$\rm \: \: \: \displaystyle\lim_{x \to 0}\rm \: \frac{ {e}^{x} - 1}{x} \: = \: 1$

$\rm \: \: \: \displaystyle\lim_{x \to 0}\rm \: \frac{ {a}^{x} - 1}{x} \: = \: loga$

Answer 2

$\rm the \: \: value \: \: of \: \: y \: \: in \: \: \boxed{ \rm\lim_{x \to - y} \: \frac{ {x}^{9} + {y}^{9} }{x + y} = 9} \: \: is \: \: \orange{\bf \pm 1}$