Question

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Answer 1

Question :-

Evaluate the following :-

$\rm \: \: \displaystyle\lim_{x \to 1}\rm \frac{sin(x - 1)}{ {x}^{2} - 1}$

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: \: \displaystyle\lim_{x \to 1}\rm \frac{sin(x - 1)}{ {x}^{2} - 1}$

If we substitute directly x = 1, then we get

$\rm \: =  \: \frac{sin(1 - 1)}{ {1}^{2} - 1}$

$\rm \: =  \: \frac{sin0}{ 1 - 1}$

$\rm \: =  \: \dfrac{0}{0}$

which is indeterminant form.

So, Consider again

$\rm \: \: \displaystyle\lim_{x \to 1}\rm \frac{sin(x - 1)}{ {x}^{2} - 1}$

can be rewritten as

$\rm \: \: \displaystyle\lim_{x \to 1}\rm \frac{sin(x - 1)}{ (x + 1)(x - 1)}$

can be further rewritten as

$\rm \: = \:\displaystyle\lim_{x \to 1}\rm \frac{1}{x + 1} \times \displaystyle\lim_{x \to 1}\rm \frac{sin(x - 1)}{ x - 1}$

$\rm \: = \: \dfrac{1}{1 + 1} \times \displaystyle\lim_{x - 1\to 0}\rm \frac{sin(x - 1)}{ x - 1}$

We know,

$\boxed{\sf{  \: \: \displaystyle\lim_{x \to 0}\rm \: \frac{sinx}{x} \: = \: 1 \: \: }}$

So, using this result, we get

$\rm \: =  \: \dfrac{1}{2} \times 1$

$\rm \: =  \: \dfrac{1}{2}$

Hence,

$\boxed{\sf{  \:\rm \: \: \displaystyle\lim_{x \to 1}\rm \frac{sin(x - 1)}{ {x}^{2} - 1} \: = \: \frac{1}{2} \: \: }}$

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Additional Information :-

$\boxed{\sf{  \: \: \displaystyle\lim_{x \to 0}\rm \: \frac{sinx}{x} \: = \: 1 \: \: }}$

$\boxed{\sf{  \: \: \displaystyle\lim_{x \to 0}\rm \: \frac{tanx}{x} \: = \: 1 \: \: }}$

$\boxed{\sf{  \: \: \displaystyle\lim_{x \to 0}\rm \: \frac{log(1 + x)}{x} \: = \: 1 \: \: }}$

$\boxed{\sf{  \: \: \displaystyle\lim_{x \to 0}\rm \: \frac{ {e}^{x} - 1}{x} \: = \: 1 \: \: }}$

$\boxed{\sf{  \: \: \displaystyle\lim_{x \to 0}\rm \: \frac{ {a}^{x} - 1}{x} \: = \: loga \: \: }}$

Answer 2

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Refer to the Attachment

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