Question

$\lim \: x \: to \: \infty \: \: ( \frac{ {x}^{2} + 1}{x + 1} - px - q) = 0$

Find the values of p and q.

Don't use L Hospital Rule​

Answer 1

Answer:

(D)

I hope this is helpful

Answer 2

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: \displaystyle\lim_{x \to \infty }\rm \: \: \bigg( \frac{ {x}^{2} + 1}{x + 1} - px - q\bigg) = 0$

can be rewritten as

$\rm \: \displaystyle\lim_{x \to \infty }\rm \: \: \bigg( \frac{ {x}^{2} + 1 - (x + 1)(px + q)}{x + 1}\bigg) = 0$

$\rm \: \displaystyle\lim_{x \to \infty }\rm \: \: \bigg( \frac{ {x}^{2} + 1 - {px}^{2} - qx - px - q }{x + 1}\bigg) = 0$

$\rm \: \displaystyle\lim_{x \to \infty }\rm \: \: \bigg( \frac{ {x}^{2}(1 - p) - x(q + p) + 1 - q }{x + 1}\bigg) = 0$

Now, as limit is given to be finite,

So,

$\rm\implies \:1 - p \: = \: 0$

$\rm\implies \:p \: = \: 1$

So, above expression can be rewritten as

$\rm \: \displaystyle\lim_{x \to \infty }\rm \: \: \bigg( \frac{ - x(q + p) + 1 - q }{x + 1}\bigg) = 0$

$\rm \: \displaystyle\lim_{x \to \infty }\rm \: \: \bigg( \frac{ x\bigg[ -(q + p) + \dfrac{1}{x} - \dfrac{q}{x}\bigg]}{x \bigg[1+ \dfrac{1}{x} \bigg]}\bigg) = 0$

$\rm \: \displaystyle\lim_{x \to \infty }\rm \: \: \bigg( \frac{ -(q + p) + \dfrac{1}{x} - \dfrac{q}{x}}{1+ \dfrac{1}{x}}\bigg) = 0$

$\rm \: - (q + p) = 0$

$\rm \: q + p = 0$

$\rm \: q + 1 = 0$

$\rm\implies \:q \: = \: - \: 1$

Hence,

$\begin{gathered}\begin{gathered}\bf\: \rm\implies \:\begin{cases} &\sf{p \: = \: 1} \\ \\ &\sf{q \: = \: - \: 1} \end{cases}\end{gathered}\end{gathered}$

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Additional Information :-

$\: \: \: \: \: \: \: \: \: \displaystyle\lim_{x \to 0}\rm \frac{sinx}{x} \: = \: 1$

$\: \: \: \: \: \: \: \: \: \displaystyle\lim_{x \to 0}\rm \frac{tanx}{x} \: = \: 1$

$\: \: \: \: \: \: \: \: \: \displaystyle\lim_{x \to 0}\rm \frac{log(1 + x)}{x} \: = \: 1$

$\: \: \: \: \: \: \: \: \: \displaystyle\lim_{x \to 0}\rm \frac{ {e}^{x} - 1}{x} \: = \: 1$

$\: \: \: \: \: \: \: \: \: \displaystyle\lim_{x \to 0}\rm \frac{ {a}^{x} - 1}{x} \: = \: loga$