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Limit of 100/(1 - x^100) - 50/(1 - x^50) as x tends to 1 | JEE Advanced

JEE Maths question with a full step-by-step solution.

Question

\displaystyle\lim_{x\to1}\left(\frac{100}{1-x^{100}}-\frac{50}{1-x^{50}}\right) = 5A

, then the value of

A

Solution

Answer: 5

Step 1: The limit is of the form

\infty-\infty

. Put

x = 1+h

h \to 0

. By the binomial theorem,

x^{n} = (1+h)^{n} = 1+nh+\frac{n(n-1)}{2}h^2+\cdots

1-x^{n} = -nh\left[1+\frac{n-1}{2}h+\cdots\right]

Step 2:

\frac{n}{1-x^{n}} = \frac{n}{-nh\left[1+\frac{n-1}{2}h+\cdots\right]} = -\frac1h\left[1-\frac{n-1}{2}h+\cdots\right] = -\frac1h+\frac{n-1}{2}+O(h)

Step 3: Putting

n = 100

and

n = 50

and subtracting,

\frac{100}{1-x^{100}}-\frac{50}{1-x^{50}} = \left(-\frac1h+\frac{99}{2}\right)-\left(-\frac1h+\frac{49}{2}\right)+O(h)

The

-\dfrac1h

terms cancel, so the limit is finite. Step 4:

\lim_{x\to1}\left(\frac{100}{1-x^{100}}-\frac{50}{1-x^{50}}\right) = \frac{99}{2}-\frac{49}{2} = \frac{50}{2} = 25

Step 5:

5A = 25 \;\Longrightarrow\; A = 5

Answer:

5.00

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