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Given :\(\lim\limits_{\text x \to \infty}\cfrac{(3\text x-1)(4\text x-2)}{(\text x+8)(\text x-1)}\)

Hence,\(\lim\limits_{\text x \to \infty}\cfrac{(3\text x-1)(4\text x-2)}{(\text x+8)(\text x-1)}\)= 12

Given:\(\lim\limits_{\text x \to \infty}\cfrac{3\text x^3-4\text x^2+6\text x-1}{2\text x^3+\text x^2-5\text x+7} \)

Hence,\(\lim\limits_{\text x \to \infty}\cfrac{3\text x^3-4\text x^2+6\text x-1}{2\text x^3+\text x^2-5\text x+7} \)=\(\cfrac32\)

As we need to find\(\lim\limits_{\text x \to0}\left\{\cfrac{3\text x^2+1}{4\text x^2-1}\right\}^{\cfrac{\text x^3}{1+\text x}} \)

lim(x→∞) {(3x2 + 1)/(4x2 - 1)}x3/(1 +x)

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.)

As it is taking indeterminate form-

∴ we need to take steps to remove this form so that we can get a finite value.

Z =\(\lim\limits_{\text x \to0}\left\{\cfrac{3\text x^2+1}{4\text x^2-1}\right\}^{\cfrac{\text x^3}{1+\text x}} \)

Take the log to bring the term in the product so that we can solve it more easily.

aking log both sides-

{using algebra of limits} Still, if we put x = ∞ we get an indeterminate form,

Take highest power of x common and try to bring x in denominator of a term so that if we put x = ∞ term reduces to 0.

{∵ log (3/4) is a negative value as 3/4 < 1}

\(\therefore\)LogeZ =

\(\Rightarrow\)Z = e-∞= 0

Hence,

\(\lim\limits_{\text x \to0}\left\{\cfrac{3\text x^2+1}{4\text x^2-1}\right\}^{\cfrac{\text x^3}{1+\text x}} \)= 0

\(\lim\limits_{\text x \to0}\cfrac{1- cos\,5\text x}{1-cos 6\text x}\)

Hence,\(\lim\limits_{\text x \to0}\cfrac{1- cos\,5\text x}{1-cos 6\text x}\)=\(\cfrac{25}{36}\)