Indeterminate Forms

Assume a function F(x)=\frac{f(x)}{g(x)} which is undefined at x=a but it may approach a limit as x approaches a. The process of determining such a limit is known as evaluation of indeterminate forms. The L’ Hospital Rule helps in the evaluation of indeterminate forms. According to this rule-
\lim_{x\to a}\frac{f(x)}{g(x)} = \lim_{x\to a}\frac{f'(x)}{g'(x)}
Provided that both f’(x) and g’(x) exist at x = a and g’(x) ≠ 0.

Types of indeterminate forms :

  1. Type \frac{0}{0}
    Suppose f(x) = 0 = g(x) as x→ a or as x→ 0
    This form can be solved directly by the application of L’ Hospital rule.
    \lim_{x\to a}\frac{f(x)}{g(x)} = \lim_{x\to a}\frac{f'(x)}{g'(x)}
    Provided that both f’(x) and g’(x) exist at x = a and g’(x) ≠ 0.
  2. Type \frac{\infty}{\infty}
    Suppose f(x) = ∞ = g(x) as x→ a or as x→ ±∞. This form can be solved by first converting it to the type \frac{0}{0} as-
    \frac{f(x)}{g(x)}=\frac{1/g(x)}{1/f(x)}
    Now we can apply L’ Hospital rule as usual to solve it. It is advised to convert to 0/0 form as the differentiation of numerator and denominator may never terminate in some problems.
  3. Type 0.\infty
    Suppose f(x) = 0 and g(x) = ∞ as x→ a or as x→ ±∞ then the product f(a).g(a) is undefined. We need to solve it by converting it to the type 0/0 or ∞/∞.
    f(x).g(x)=\frac{f(x)}{1/g(x)} or \frac{g(x)}{1/f(x)}
    Now we need to apply L’ Hospital rule.
  4. Type \infty - \infty
    Suppose f(x) = ∞ = g(x) as x→ a. this type is solved by again converting to the 0/0 form by following method :
    f(x)-g(x)=\frac{\frac{1}{g(x)}-\frac{1}{f(x)}}{\frac{1}{f(x)g(x)}}
    As we achieve 0/0 form, now we can apply L’ Hospital rule.
  5. Type 0^0, \infty^0, 1^{\infty}
    To evaluate these forms consider:
    y(x)=f(x)^{g(x)}
    Taking logarithm both sides
    \lny=g(x)\lnf(x)
    Taking the limit as x→ a or x→ ±∞
    \lim_{x\to a}\lny=k
    Then k=\lim_{x\to a}(\lny)=\ln(\lim_{x\to a}y)
    =\ln(\lim_{x\to a} f(x)^{g(x)})
    \lim_{x\to a}f(x)^{g(x)}=e^k

Note –
If f’(x) and g’(x) do not exist at x=a then we need to perform the differentiation again until the derivatives of f(x) and g(x) become valid.



Example-1:
Evaluate \lim_{x\to 1}\frac{1+\lnx-x}{1-2x+x^2}



Explanation :
As the given function assumes 0/0 form at x = 1, so we can directly apply L’ Hospital rule.
F(x)=\frac{f(x)}{g(x)}=\frac{f'(x)}{g'(x)}
f(x)=1+\lnx-x \scriptstyle\implies f'(x)=0+1/x-1
g(x)=1-2x+x^2 \scriptstyle\implies g'(x)=-2+2x
This forms 0/0 form again. Hence we apply L’ Hospital rule again.
f''(x)=\frac{-1}{x^2} and g''(x)=2
Thus \lim_{x\to 1} F(x)=\frac{f''(x)}{g''(x)}
=\lim_{x \to 1}\frac{-1/x^2}{2}=\frac{-1}{2}



Example-2:
Evaluate \lim_{x\to 1}\log(1-x).\cot\frac{\pi x}{2}

Explanation :
The given function assumes 0.∞ form. We will first rewrite it in \frac{\infty}{\infty} form.
\lim_{x\to 1}log(1-x).cot\frac{\pi x}{2}=\lim_{x\to 1}\frac{\ln(1-x)}{\tan \pi x/2}
Now we apply L’ Hospital rule to get
\lim_{x\to 1}\frac{1/(1-x).(-1)}{\pi/2.\sec^2 \pi x/2}
This forms \frac{\infty}{\infty} form again. We rewrite it in 0/0 form as-
\lim_{x \to 1}\frac{-2\cos^2\pi x/2}{\pi (1-x)}
Now apply L’ Hospital rule again.
\Lim_{x \to 1}\frac{-2}{\pi}.\frac{2.\cos\pi x/2.(-\sin \pi x/2).\pi/2}{-1}=0

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