Limits, Continuity and Differentiability
79570
If \(\lim _{x \rightarrow 1} \frac{x^{2}-a x+b}{(x-1)}=5\), then \((a+b)\) is equal to
Explanation:
(C) : We have,
\(\lim _{x \rightarrow 1} \frac{x^{2}-a x+b}{x-1}=5 \tag{i}\)
On applying L- Hospital rule, we get -
\(\lim _{x \rightarrow 1} \frac{2 x-a}{1}=5 \)
\(\text { at } \quad \mathrm{x}=1 \)
\(2 \times 1-a=5 \)
\(a=2-5 \)
\(a=-3 \)
Value \(a=-3\) substituting equation (i)-
\(\lim _{x \rightarrow 1} \frac{x^{2}-(-3 x)+b}{x-1}=5\)
\(1+3+b=0\)
\(b=-4\)
\((\mathrm{a}+\mathrm{b})=-3-4=-7\)