Question

$\underset{x\to 1}{\mathop{\lim }}\,\dfrac{x+{{x}^{2}}+....+{{x}^{n}}-n}{x-1}$ equals:
(a) n
(b) 0
(c) $\dfrac{{{n}^{2}}}{2}$
(d) $\dfrac{n\left( n+1 \right)}{2}$

Answer 1

Hint: From different forms of limit we can find that the above given one is an indefinite form. So, we need to use the L' Hospital's rule to simplify it further.

Complete step-by-step answer:

$\underset{x\to a}{\mathop{\lim }}\,\dfrac{f\left( x \right)}{g\left( x \right)}=\underset{x\to a}{\mathop{\lim }}\,\dfrac{f'\left( x \right)}{g'\left( x \right)}$

Let us look at some of the definitions and forms of limits.

LIMIT: Let y = f(x) be a function of x. If at x = a, f(x) takes indeterminate form, then we consider the value of the function which is very near to a. If these values tend to a definite unique number as x tends to a, then the unique number, so obtained is called the limit of f(x) at x = a and we write it as $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)$.

LEFT HAND AND RIGHT HAND LIMITS:

If the values of the function at the points which are very near to a on the left tends to a definite unique number as x tends to a, then the unique number, so obtained, is called the left hand limit of f(x) at x = a. We write it as:

$f\left( a-0 \right)=\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,f\left( a-h \right)$

Similarly, right hand limit can be written as:

$f\left( a+0 \right)=\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,f\left( a+h \right)$

Existence of limit: $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)$ exists, if

$\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ and $$\underset{x\to

{{a}^{+}}}{\mathop{\lim }},f\left( x \right)$$ both exist.

{{a}^{+}}}{\mathop{\lim }}\,f\left( x \right)$$ Uniqueness of limit: If $$\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)$$ exists, then it is unique. There cannot be two distinct numbers l and m such that when x tends to a, then the function f(x) tends to both l and m. INDETERMINATE FORMS: If direct substitution of x = a while evaluating $$\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)$$ leads to one of the following form: $$\dfrac{0}{0};\dfrac{\infty }{\infty };\infty -\infty ;0\times \infty ;{{1}^{\infty }};{{0}^{0}};{{\infty }^{0}}$$ Then it is called the indeterminate form. These limits can be evaluated by using L' Hospital's rule or some other methods. L' Hospital's Rule: If f(x) and g(x) be two functions of x such that $$\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to a}{\mathop{\lim }}\,g\left( x \right)=0$$ Both are continuous at x = a Both are differentiable at x = a f'(x) and g'(x) are continuous at the point x = a, then $$\underset{x\to a}{\mathop{\lim }}\,\dfrac{f\left( x \right)}{g\left( x \right)}=\underset{x\to a}{\mathop{\lim }}\,\dfrac{f'\left( x \right)}{g'\left( x \right)}$$ Now, from the given equation in the question we have, $$\Rightarrow \underset{x\to 1}{\mathop{\lim }}\,\dfrac{x+{{x}^{2}}+....+{{x}^{n}}-n}{x-1}$$ Here, as x tends to 1 then both the numerator and denominator will be 0. Thus, we can apply the L' Hospital's Rule. Now, on differentiating the numerator and denominator we get, $$\Rightarrow \underset{x\to 1}{\mathop{\lim }}\,\dfrac{1+2x+....+n{{x}^{n-1}}}{1}$$ As we already know that differentiation of constant is 0. $$\left[ \because \dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}} \right]$$ Now, on applying the limit we get, $$\begin{aligned} & \Rightarrow 1+2\left( 1 \right)+....+n{{\left( 1 \right)}^{n-1}} \\\ & \Rightarrow 1+2+.....+n \\\ & \Rightarrow \dfrac{n\left( n+1 \right)}{2} \\\ \end{aligned}$$ Hence, the correct option is (d). Note: Instead of solving the above indeterminate form using the L' Hospital's Rule we can also solve it by using the methods like limit by factorisation and limit by substitution. Whereas, simplifying it using the L' Hospital's Rule is easy and less confusing here. Both the methods give the same result. It is important to note that applying the limit to the above function gives an indeterminate and then to be solved accordingly. While differentiating the terms we need to be careful about the result of the differentiation and should not neglect any of the terms because neglecting any one of the terms changes the result completely.