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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
If limx1(1001x100501x50)=5A\displaystyle\lim_{x\to1}\left(\frac{100}{1-x^{100}}-\frac{50}{1-x^{50}}\right) = 5A, then the value of AA.
Solution
Answer: 5
Step 1: The limit is of the form \infty-\infty. Put x=1+hx = 1+h, h0h \to 0. By the binomial theorem,
xn=(1+h)n=1+nh+n(n1)2h2+x^{n} = (1+h)^{n} = 1+nh+\frac{n(n-1)}{2}h^2+\cdots
1xn=nh[1+n12h+]1-x^{n} = -nh\left[1+\frac{n-1}{2}h+\cdots\right]
Step 2:
n1xn=nnh[1+n12h+]=1h[1n12h+]=1h+n12+O(h)\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=100n = 100 and n=50n = 50 and subtracting,
1001x100501x50=(1h+992)(1h+492)+O(h)\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 1h-\dfrac1h terms cancel, so the limit is finite. Step 4:
limx1(1001x100501x50)=992492=502=25\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    A=55A = 25 \;\Longrightarrow\; A = 5
Answer: 5.005.00
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