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Given,

\(\lim\limits_{\text x \to a}\cfrac{\text x^3-a^3}{\text x-a}=\lim\limits_{\text a\to1}\cfrac{\text x^4-1}{\text x-1} \)

Note: To solve the problems of limit similar to one in our question we use the formula mentioned below which can be derived using binomial theorem.

Formula to be used:\(\lim\limits_{\text x \to a}\cfrac{(\text x)^n-(a)^n}{\text x-a} \)= nan -1

Using the formula we have –

3a3–1 = 4(1)4–1

⇒ 3a2 = 4

⇒ a2 = 4/3

∴ a = ± (2/√3)

As we need to find\(\lim\limits_{\text x \to a}\left\{\cfrac{sin\,\text x}{sin\, a}\right\}^{\cfrac{1}{\text x-a}} \)

lim(x→0) {sin x/sin a}1/(x - a)

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 =\(\lim\limits_{\text x \to a}\left\{\cfrac{sin\,\text x}{sin\, a}\right\}^{\cfrac{1}{\text x-a}} \)

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

Taking log both sides-

{∵ log am = m log a}

Now it gives us a form that can be reduced to\(\lim\limits_{\text x \to0}\cfrac{log(1+\text x)}{\text x}=1\)

Dividing numerator and denominator by\(\cfrac{sin\,\text x-sin\,a}{sin\,a}\)to get the desired form and using algebra of limits we have-

Now it gives us a form that can be reduced to\(\lim\limits_{\text x \to0}\cfrac{sin\,\text x}{\text x}=1\)

Try to use it. We are basically proceeding with a hit and trial attempt.

∵ sin (A + B) = sin A cos B + cos A sin B

Use the formula-\(\lim\limits_{\text x \to0}\cfrac{sin\,\text x}{\text x}=1\)

⇒ log Z = cot a – 0

∴ log Z = cot a

∴ Z = ecot a

Hence,

\(\lim\limits_{\text x \to a}\left\{\cfrac{sin\,\text x}{sin\, a}\right\}^{\cfrac{1}{\text x-a}} \)= ecot a

To evaluate:\(\lim\limits_{\text x \to2} \)(5 -x)

lim(x→0)

Formula used:

We have,

Thus, the value of\(\lim\limits_{\text x \to2} \)(5 -x) is 3.