Question

limit x tends to 1 (2/1-x² + 1/x-1)

By factorization method
Limits and derivative
Class 11 maths​

Answer 1

Answer:

That's it my friend.

BTW,same class's student here......:)

Answer 2

The limit of the expression $(2/(1-x^2) + 1/(x-1))$as x approaches 1 is 6

To find the limit of the expression $(2/(1-x^2) + 1/(x-1))$ as x approaches 1, we can use the factorization method.

First, we can factor the numerator and denominator of the first fraction:

$2/(1-x^2) = 2/(1-x)(1+x)$

Then we can factor the numerator and denominator of the second fraction:

$1/(x-1) = 1/(x-1)(1/(x-1))$

Now, we can substitute the factored forms of the fractions into the original expression:

$(2/(1-x)(1+x) + 1/(x-1)(1/(x-1))$

We can now substitute x=1 into the expression above:

$(2/(1-1)(1+1) + 1/(1-1)(1/(1-1))$

This gives us:

$(2/(0)(2) + 1/(0)(1))$

Now we can simplify it further:

$(2/0 + 1/0)$

This is an indeterminate form since the denominator is zero, in such cases we can use L'Hopital's rule.

L'Hopital's rule states that if we have an expression in the form f(x)/g(x) and both f(x) and g(x) approach zero or infinity as x approaches a particular value, we can find the limit of the expression by taking the derivative of f(x) and g(x) and finding the limit of the ratio of their derivatives.

The derivative of the numerator is $2(1+x) + (x-1)^-2$ and the derivative of the denominator is $(1-x^2)^-1 + (x-1)^-2$

The limit of the ratio of the derivatives as x tends to 1 is $2(1+1) + (1-1)^-2 + (1-1)^-2 = 4 + 1 + 1 = 6$

Therefore, the limit of the expression$(2/(1-x^2) + 1/(x-1))$ as x approaches 1 is 6

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