As we need to find \(\lim\limits_{\text x \to1}\left\{\cfrac{\text x^3+2\text x^2+\text x+1}{\text x^2+2\text x+3}\right\}^{\cfrac{1-cos(\text x-1)}{(\text x-1)^2}} \)
lim(x→1) {(x^3 + 2x^2 + x + 1)/(x^2 + 2x + 3)}^{1 - cos(x - 1)/(x - 1)^2}
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞,1∞ .. etc.)
Let Z = \(\lim\limits_{\text x \to1}\left\{\cfrac{\text x^3+2\text x^2+\text x+1}{\text x^2+2\text x+3}\right\}^{\cfrac{1-cos(\text x-1)}{(\text x-1)^2}} \)
\(=\left(\cfrac56\right)^{\frac00}\)(indeterminate)
As it is taking indeterminate form-
∴ we need to take steps to remove this form so that we can get a finite value.
Z = \(\lim\limits_{\text x \to1}\left\{\cfrac{\text x^3+2\text x^2+\text x+1}{\text x^2+2\text x+3}\right\}^{\cfrac{1-cos(\text x-1)}{(\text x-1)^2}} \)
Take the log to bring the power term in the product so that we can solve it more easily.
Taking log both sides-
log Z = \(\lim\limits_{\text x \to1}\left\{\cfrac{\text x^3+2\text x^2+\text x+1}{\text x^2+2\text x+3}\right\}^{\cfrac{1-cos(\text x-1)}{(\text x-1)^2}} \)
{∵ log am = m log a}
using algebra of limits-
Hence,
\(\lim\limits_{\text x \to1}\left\{\cfrac{\text x^3+2\text x^2+\text x+1}{\text x^2+2\text x+3}\right\}^{\cfrac{1-cos(\text x-1)}{(\text x-1)^2}} \) = \(\sqrt{\cfrac56}\)
