Question

Find: -
\[\LARGE{\lim_{x \rightarrow 1}\frac{x^m-1}{x^n-1}}=?\]
m & n are integers.

Answer 1

Answer is
m/n.

But how to get it ?
Please help:)

Answer 2

@UnkleRhaukus :)

Answer 3

l'Hôpital's rule!

Answer 4

well use this important identity\[x^k-1=(x-1)(x^{k-1}+x^{k-2}+...+x+1)\]

Answer 5

\[\large{x^m -1 = (x-1)(x^{m-1}+x^{m-2}+..+x+1)}\]
\[\large{x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x+1)}\]

divide them

Answer 6

x-1 will cancel out...

and then there will be work of LIMITS..

Answer 7

Let f(x)=x^m-1 then find f'(x)
Let g(x)=x^n-1 then fing g'(x)

Answer 8

\[\large{lim_{x \rightarrow 1}\frac{x^{m-1}+x^{m-2}+..+x+1}{x^{n-1}+x^{n-2}+..+x+1}}\]
\[\large{\frac{1+1+..+1+1}{1+1+...1+1}}\]
\[\large{\frac{m}{n}}\]

Answer 9

in the numerator it will be :

m times 1

and in the denominator it will be :

n times 1

hence we have m/n as the answer

Answer 10

Now, f'(x)=mx^(m-1) and g'(x)=nx^(n-1)
Now,