Question

Solve ques no 6

Show that lim x to 1 (x+x^2+x^3+—-x^n)/(x-1) = n(n+1)/2

Answer 1

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

Consider

$\displaystyle\lim_{x \to 1}\rm \frac{(x + {x}^{2} + {x}^{3} + - - - {x}^{n}) - n}{x - 1}$

If we substitute directly x = 1, we get

$\rm \: = \: \dfrac{ \underbrace{1 + 1 + 1 + - - - + 1} - n}{1 - 1}$

$\rm \: = \: \dfrac{n - n}{0}$

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

which is indeterminant form.

So, Consider again

$\displaystyle\lim_{x \to 1}\rm \frac{(x + {x}^{2} + {x}^{3} + - - - {x}^{n}) - n}{x - 1}$

Now,

n can be rewritten as

$\rm \: n = \underbrace{1 + 1 + 1 + - - - + 1} \\ \rm \: \: \: \: \: \: \: \: n \: times$

So, above expression can be rewritten as

$\rm \: = \: \displaystyle\lim_{x \to 1}\rm \frac{(x + {x}^{2} + {x}^{3} + - - - {x}^{n}) - (1 + 1 + 1 + - - + 1)}{x - 1}$

$\rm \: = \: \displaystyle\lim_{x \to 1}\rm \frac{(x - 1) + ({x}^{2} - 1) +({x}^{3} - 1)+ - - - ({x}^{n} - 1)}{x - 1}$

can be further rewritten as

$\rm \: = \: \displaystyle\lim_{x \to 1}\rm \frac{x - 1}{x - 1} + \displaystyle\lim_{x \to 1}\rm \frac{ {x}^{2} - 1}{x - 1} + \displaystyle\lim_{x \to 1}\rm \frac{ {x}^{3} - 1}{x - 1} + - - - + \displaystyle\lim_{x \to 1}\rm \frac{ {x}^{n} - 1}{x - 1}$

can be rewritten as

$\rm \: = \: 1 + \displaystyle\lim_{x \to 1}\rm \frac{ {x}^{2} - {1}^{2} }{x - 1} + \displaystyle\lim_{x \to 1}\rm \frac{ {x}^{3} - {1}^{3} }{x - 1} + - - - + \displaystyle\lim_{x \to 1}\rm \frac{ {x}^{n} - {1}^{n} }{x - 1}$

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

$\rm \: = \: 1 + 2 {(1)}^{2 - 1} + 3 {(1)}^{3 - 1} + - - - + n{(1)}^{n - 1}$

$\rm \: = \: 1 + 2 + 3 + - - - - + n$

$\rm \: = \: \dfrac{n(n + 1)}{2}$

Hence,

$\boxed{\tt{ \displaystyle\lim_{x \to 1}\rm \frac{(x + {x}^{2} + {x}^{3} + - - - {x}^{n}) - n}{x - 1} = \frac{n(n + 1)}{2} }}$

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ADDITIONAL INFORMATION

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

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

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

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

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

Answer 2

Now, Let us consider.

$\rm \lim_{(x \to1)} \frac{(x + {x}^{2} + {x}^{3} + - - - {x}^{n} ) - n}{x - 1}$

If we substitute x = 1 directly, we obtain.

$\rm = \frac{1 + 1 + 1 + - - + 1 - n}{1 - 1}$

$\rm = \frac{0}{0}$

⬟ It is in Indeterminant form.

Let us consider,

$\rm \lim_{(x \to1)} \frac{(x + {x}^{2} + {x}^{3} + - - - {x}^{n} ) - n}{x - 1}$

Now, n can be also rewritten as.

$\rm n = \underbrace{1 + 1 + 1 + - - - + 1} \\ \rm n \: \: times$

so, The above Considered Expression can be rewritten as,

$\rm⇒ \lim_{(x \to1)} \frac{(x + {x}^{2} + {x}^{3} + - - - {x}^{n} ) - (1 + 1 + 1 + - - - + 1)}{x - 1}$

$\rm⇒\rm \lim_{(x \to1)} \frac{(x - 1) + ( {x}^{2} - 1) + ( {x}^{3} - 1) + - - - + ( {x}^{n} - 1)}{x - 1}$

Finally, by using

$\boxed{ \boxed{ \bf \lim_{x \to a} \: \frac{ {x}^{n} - {a}^{n} }{x - a} = n {a}^{n - 1} }}$

we obtain.

$\rm⇒ 1 + 2 {(1)}^{2 - 1} + 3 {(1)}^{3 - 1} + - - - + n {(1)}^{n - 1}$

$\rm⇒1 + 2 + 3 + - - - + n$

$\rm⇒ \frac{n(n + 1)}{2}$

$\orange{▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄}$

$\boxed{ \boxed{ \therefore \color{green}{\rm \lim_{(x \to1)} \frac{(x + {x}^{2} + {x}^{3} + - - - {x}^{n} ) - n}{x - 1} = \frac{n(n + 1)}{2} }}}$

Hence, Proved