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Question: How do you find the limit of \[\dfrac{{{x}^{n}}-1}{x-1}\] as \[x\] approaches 1?...

How do you find the limit of xn−1x−1\dfrac{{{x}^{n}}-1}{x-1} as xx approaches 1?

Explanation

Solution

We can solve this question using simple calculations and formulas. Here we have to find the value within the limit 1 as it is the restriction given in the question. We will expand the numerator in such a way we simplify the equation. After it gets simplified we should apply limits to the simplified equation to get the result.

Complete step by step answer:
Given equation can be represented as
lim⁡x→1xn−1x−1\displaystyle \lim_{x \to 1}\dfrac{{{x}^{n}}-1}{x-1}
To solve the above equation we have to rewrite the numerator as
⇒(x−1)(xn−1+xn−2+.....+x+1)\Rightarrow \left( x-1 \right)\left( {{x}^{n-1}}+{{x}^{n-2}}+.....+x+1 \right)
We had written the above equation from numerator using this formula
an−bn=(a−b)(an−1+ban−2+b2an−3+.....+bn−2a+bn−1){{a}^{n}}-{{b}^{n}}=\left( a-b \right)\left( {{a}^{n-1}}+b{{a}^{n-2}}+{{b}^{2}}{{a}^{n-3}}+.....+{{b}^{n-2}}a+{{b}^{n-1}} \right)
So in our case numerator is xn−1{{x}^{n}}-1 so we can write it as xn−1n{{x}^{n}}-{{1}^{n}}
By substituting in above formula we get
⇒(x−1)(xn−1+xn−2+.....+x+1)\Rightarrow \left( x-1 \right)\left( {{x}^{n-1}}+{{x}^{n-2}}+.....+x+1 \right)
Now the equation will become
⇒lim⁡x→1(x−1)(xn−1+xn−2+.....+x+1)x−1\Rightarrow \displaystyle \lim_{x \to 1}\dfrac{\left( x-1 \right)\left( {{x}^{n-1}}+{{x}^{n-2}}+.....+x+1 \right)}{x-1}
Here we can cancel x−1x-1 in both numerator and denominator. so the new equation we will get is
⇒lim⁡x→1(xn−1+xn−1+xn−2+.....+x+1)\Rightarrow \displaystyle \lim_{x \to 1}\left( {{x}^{n-1}}+{{x}^{n-1}}+{{x}^{n-2}}+.....+x+1 \right)
As we can see the above equation is a summation of terms from 1 to xn−1{{x}^{n-1}} terms . so we can represent the equation as
⇒lim⁡x→1∑i=0n−1xi\Rightarrow \displaystyle \lim_{x \to 1}\sum\limits_{i=0}^{n-1}{{{x}^{i}}}
Now we have to apply limits to the function we have. Substituting 1 in place of x.
⇒∑i=0n−11i\Rightarrow \sum\limits_{i=0}^{n-1}{{{1}^{i}}}
Now we have to apply summation using its limits. But 1i{{1}^{i}} is always 1 so we can say the answer is the difference between the limits.
⇒10+11+.......1n−1\Rightarrow {{1}^{0}}+{{1}^{1}}+{{.......1}^{n-1}}
From there we can say we have added 1 for nn times.
So the solution is nn

lim⁡x→1xn−1x−1=n\displaystyle \lim_{x \to 1}\dfrac{{{x}^{n}}-1}{x-1}=n

Note: We can also solve this equation using L'Hospital's' Rule. We must be careful while expanding or merging the equations if anything went wrong in those steps then the whole sum will go wrong. There are many ways to simplify the equations formed by the above method, one among them ,you can use your own simplification method.