Factorize both the numerator and denominator:
\( x^2 - 9 = (x - 3)(x + 3), \)
\( x^2 - 4x + 3 = (x - 3)(x - 1). \)
So, the expression becomes:
\( \frac{(x - 3)(x + 3)}{(x - 3)(x - 1)}. \)
Cancel out the \((x - 3)\) terms:
\( \frac{x + 3}{x - 1}. \)
Substitute \(x = 3\):
\( \frac{3 + 3}{3 - 1} = \frac{6}{2} = 3. \)
LIST I | LIST II | ||
A. | \(\lim\limits_{x\rightarrow0}(1+sinx)^{2\cot x}\) | I. | e-1/6 |
B. | \(\lim\limits_{x\rightarrow0}e^x-(1+x)/x^2\) | II. | e |
C. | \(\lim\limits_{x\rightarrow0}(\frac{sinx}{x})^{1/x^2}\) | III. | e2 |
D. | \(\lim\limits_{x\rightarrow\infty}(\frac{x+2}{x+1})^{x+3}\) | IV. | Β½ |