Math, asked by vaibhavgowda10, 11 months ago

evaluate Lim↓x-1(x^15-1/x^10-1)​

Answers

Answered by srijadutta2014
12

Answer:

Step-by-step explanation:

Attachments:
Answered by Anonymous
7

Limits

We are asked to evaluate the following limits:

\longrightarrow \lim \limits_{x \to 1} \dfrac{x^{15} - 1}{x^{10} - 1}

By directly substituting the limits x = 1, we get:

\dfrac{1^{15} - 1}{1^{10} - 1} = \dfrac{1 - 1}{1 - 1} = \boxed{\dfrac{0}{0}}

Which is an indeterminate form therefore we have to use another method to solve the problem.

Solution:

We can see that the factorisation is not possible. So, dividing numerator and denominator by x - 1.

\implies \lim \limits_{x \to 1} \dfrac{\frac{x^{15} - 1}{x - 1}}{\frac{x^{10} - 1}{x - 1}}

\implies \lim \limits_{x \to 1} \Bigg\{\bigg(\dfrac{x^{15} - 1}{x - 1}\bigg) \div \bigg(\dfrac{x^{10} - 1}{x - 1}\bigg)\Bigg\}

\implies \lim \limits_{x \to 1} \bigg(\dfrac{x^{15} - 1^{10}}{x - 1}\bigg) \div  \lim \limits_{x \to 1} \bigg(\dfrac{x^{10} - 1^{10}}{x - 1}\bigg)

Now both the limits are in the form of \lim \limits_{x \to a} \dfrac{x^n - a^n}{x - a} = na^{n-1} so applying this formula in it, we get:

\implies \dfrac{15(1)^{15 - 1}}{10(1)^{10 - 1}}

\implies \dfrac{15(1)^{14}}{10(1)^{9}}

\implies \dfrac{15}{10}

\implies \dfrac{3}{2}

Therefore the required answer is:

\boxed{\lim \limits_{x \to 1} \dfrac{x^{15} - 1}{x^{10} - 1} = \dfrac{3}{2}}

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