Lim X → a Sin √ X − Sin √ a X − a - Mathematics

प्रश्न

$lim_{x \to a} \frac{\sin \sqrt{x} - \sin \sqrt{a}}{x - a}$

उत्तर

$lim_{x \to a} \frac{\sin \sqrt{x} - \sin \sqrt{a}}{x - a}$
$= lim_{x \to a} \frac{2 \cos (\frac{\sqrt{x} + \sqrt{a}}{2}) \sin (\frac{\sqrt{x} - \sqrt{a}}{2})}{(\sqrt{x} + \sqrt{a}) (\sqrt{x} - \sqrt{a})}$
$= lim_{x \to a} \frac{2 \cos (\frac{\sqrt{x} + \sqrt{a}}{2})}{\sqrt{x} + \sqrt{a}} \times \frac{\sin (\frac{\sqrt{x} - \sqrt{a}}{2})}{2 (\frac{\sqrt{x} - \sqrt{a}}{2})}$
$= \frac{1}{\sqrt{a} + \sqrt{a}} \cos (\frac{2 \sqrt{a}}{2})$
$\Rightarrow \frac{1}{2 \sqrt{a}} \cos \sqrt{a}$

shaalaa.com

Concept of Limits - Algebra of Limits

या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?

Q 17 Q 16 Q 18

पाठ 29: Limits - Exercise 29.8 [पृष्ठ ६२]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11

पाठ 29 Limits
Exercise 29.8 | Q 17 | पृष्ठ ६२

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

view
व्हिडिओ ट्यूटोरियल For All Subjects
Concept of Limits - Algebra of Limits

video tutorial00:08:11

संबंधित प्रश्‍न

Show that $lim_{x \to 0} \frac{x}{| x |}$ does not exist.

$lim_{x \to 0} 9$

$lim_{x \to - 1} \frac{x^{3} + 1}{x + 1}$

$lim_{x \to \sqrt{3}} \frac{x^{2} - 3}{x^{2} + 3 \sqrt{3} x - 12}$

$lim_{x \to 0} \frac{{(a + x)}^{2} - a^{2}}{x}$

$lim_{x \to 1} \frac{1 - x^{- 1 / 3}}{1 - x^{- 2 / 3}}$

$lim_{x \to 2} [\frac{1}{x - 2} - \frac{2 (2 x - 3)}{x^{3} - 3 x^{2} + 2 x}]$

$lim_{x \to 27} \frac{(x^{1 / 3} + 3) (x^{1 / 3} - 3)}{x - 27}$

If $lim_{x \to 3} \frac{x^{n} - 3^{n}}{x - 3} = 108,$ find the value of n.

$lim_{x \to \infty} \frac{5 x^{3} - 6}{\sqrt{9 + 4 x^{6}}}$

$lim_{x \to \infty} \sqrt{x^{2} + c x - x}$

$lim_{x \to \infty} [\frac{x^{4} + 7 x^{3} + 46 x + a}{x^{4} + 6}]$ where a is a non-zero real number.

$lim_{x \to 0} [\frac{x^{2}}{\sin x^{2}}]$

$lim_{x \to 0} \frac{\tan m x}{\tan n x}$

$lim_{x \to 0} \frac{7 x \cos x - 3 \sin x}{4 x + \tan x}$

$lim_{x \to 0} \frac{\cos a x - \cos b x}{\cos c x - \cos d x}$

$lim_{θ} \to 0 \frac{\sin 3 θ}{\tan 2 θ}$

$lim_{x \to 0} \frac{2 \sin x^{\circ} - \sin 2 x^{\circ}}{x^{3}}$

$lim_{x \to 0} \frac{x \cos x + \sin x}{x^{2} + \tan x}$

Evaluate the following limits:

$lim_{x \to 0} \frac{2 \sin x - \sin 2 x}{x^{3}}$

$lim_{x \to 0} (c o s e c x - \cot x)$

$lim_{x \to \frac{π}{4}} \frac{1 - \sin 2 x}{1 + \cos 4 x}$

$lim_{x \to π} \frac{1 + \cos x}{\tan^{2} x}$

$lim_{n \to \infty} {(1 + \frac{x}{n})}^{n}$

Write the value of $lim_{x \to 0^{-}} [x] .$

$lim_{x \to \infty} {\frac{3 x^{2} + 1}{4 x^{2} - 1}}^{\frac{x^{3}}{1 + x}}$

$lim_{x \to 0} \frac{\sqrt{1 - \cos 2 x}}{x} .$

Write the value of $lim_{x \to 0^{-}} [x] .$

If $f (x) = x \sin (1 / x), x \neq 0,$ then $lim_{x \to 0} f (x) =$

$lim_{h \to 0} {\frac{1}{h \sqrt[3]{8 + h}} - \frac{1}{2 h}} =$

$lim_{x \to 3} \frac{\sum_{r = 1}^{n} x^{r} - \sum_{r = 1}^{n} 3^{r}}{x - 3}$ is real to

$lim_{n \to \infty} \frac{1 - 2 + 3 - 4 + 5 - 6 + . . . . + (2 n - 1) - 2 n}{\sqrt{n^{2} + 1} + \sqrt{n^{2} - 1}}$ is equal to

$lim_{x \to π / 4} \frac{4 \sqrt{2} - {(\cos x + \sin x)}^{5}}{1 - \sin 2 x}$ is equal to

The value of $lim_{x \to π / 2} (\sec x - \tan x)$ is

$lim_{x \to \infty} \frac{| x |}{x}$ is equal to

Evaluate the following Limit:

$lim_{x \to 0} \frac{{(1 + x)}^{n} - 1}{x}$

Evaluate: $lim_{x \to 1} \frac{{(1 + x)}^{6} - 1}{{(1 + x)}^{2} - 1}$

Evaluate the following limits: $lim_{x \to 3} [\frac{\sqrt{x + 6}}{x}]$

Evaluate the following limits: $lim_{x \to 5} [\frac{x^{3} - 125}{x^{5} - 3125}]$

Evaluate the following limit:

$lim_{x \to 7} [\frac{(\sqrt[3]{x} - \sqrt[3]{7}) (\sqrt[3]{x} + \sqrt[3]{7})}{x - 7}]$

Lim X → a Sin √ X − Sin √ a X − a - Mathematics

Advertisements

Advertisements

प्रश्न

उत्तर

APPEARS IN

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

Select a course

Lim X → a Sin √ X − Sin √ a X − a - Mathematics

Advertisements

Advertisements

प्रश्न

उत्तरShow Solution

APPEARS IN

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

Select a course

उत्तर