Limit and Continuity MCQ Quiz in मराठी - Objective Question with Answer for Limit and Continuity - मोफत PDF डाउनलोड करा

गणना:

दिलेले आहे, $\underset{x\to 0^{+}}{lim}{(\sin (kx)+\cos (kx)+x)}^{\frac{3}{x}}$ = e 9

समजा, L = $\underset{x\to 0^{+}}{lim}\frac{3}{x}(\sin (kx)+\cos (kx)+x-1)$

= $\underset{x\to 0^{+}}{lim}\frac{3}{x}(\sin (kx)+\cos (kx)+x-1)$

= $\underset{x\to 0^{+}}{lim}3(\frac{\sin (kx)}{x}+1-\frac{1-\cos (kx)}{x})$

= $\underset{x\to 0^{+}}{lim}3(\frac{\sin (kx)}{(kx)}\cdot k+1-\frac{2{\sin }^2\frac{kx}{2}}{\frac{k^2{x}^2}{4}}\cdot \frac{k^{2}x}{4})$

= $\underset{x\to 0^{+}}{lim}3(\frac{\sin (kx)}{(kx)}\cdot k+1-2\left(\frac{\sin \frac{kx}{2}}{\frac{kx}{2}}\right)\cdot \frac{k^{2}x}{4})$

= $\underset{x\to 0^{+}}{lim}3(k+1-\frac{k^{2}x}{2})$

= ३(के + १)

∴ $\underset{x\to 0^{+}}{lim}{(\sin (kx)+\cos (kx)+x)}^{\frac{3}{x}}$ = $e^{\frac{3}{x}(\underset{x\to 0^{+}}{lim}\sin (kx)+\cos (kx)+x-1)}$

= e3(k + 1) = e9 

⇒ 3(k + 1) = 9

⇒ k + 1 = 3

⇒ k = 2

∴ k चे मूल्य 2 आहे.

पर्याय 1 योग्य आहे.

गणना:

दिलेले आहे, $\underset{x\to 0^{+}}{lim}{(\sin (kx)+\cos (kx)+x)}^{\frac{3}{x}}$ = e 9

समजा, L = $\underset{x\to 0^{+}}{lim}\frac{3}{x}(\sin (kx)+\cos (kx)+x-1)$

= $\underset{x\to 0^{+}}{lim}\frac{3}{x}(\sin (kx)+\cos (kx)+x-1)$

= $\underset{x\to 0^{+}}{lim}3(\frac{\sin (kx)}{x}+1-\frac{1-\cos (kx)}{x})$

= $\underset{x\to 0^{+}}{lim}3(\frac{\sin (kx)}{(kx)}\cdot k+1-\frac{2{\sin }^2\frac{kx}{2}}{\frac{k^2{x}^2}{4}}\cdot \frac{k^{2}x}{4})$

= $\underset{x\to 0^{+}}{lim}3(\frac{\sin (kx)}{(kx)}\cdot k+1-2\left(\frac{\sin \frac{kx}{2}}{\frac{kx}{2}}\right)\cdot \frac{k^{2}x}{4})$

= $\underset{x\to 0^{+}}{lim}3(k+1-\frac{k^{2}x}{2})$

= ३(के + १)

∴ $\underset{x\to 0^{+}}{lim}{(\sin (kx)+\cos (kx)+x)}^{\frac{3}{x}}$ = $e^{\frac{3}{x}(\underset{x\to 0^{+}}{lim}\sin (kx)+\cos (kx)+x-1)}$

= e3(k + 1) = e9 

⇒ 3(k + 1) = 9

⇒ k + 1 = 3

⇒ k = 2

∴ k चे मूल्य 2 आहे.

पर्याय 1 योग्य आहे.